Chuyên đề 6: Hệ phương trình (có đáp án)

$\left\{\begin{array}{l}2x+3y=7\\ x-y=1\end{array}\right.\iff \left\{\begin{array}{l}2x+3y=7\\ 3x-3y=3\end{array}\right.\iff \left\{\begin{array}{l}5x=10\\ 2x+3y=7\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ 2.2+3y=7\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$

$\left\{\begin{array}{l}2x-y=3\\ 3x+2y=8\end{array}\right.\iff \left\{\begin{array}{l}7x=14\\ 2x-y=3\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ 4-y=3\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=1\end{array}\right..$

Vậy hệ phương trình có nghiệm duy nhất là (x;y)=(2;1)

Hệ phương trình

$\left\{\begin{array}{l}3x+y=10\\ 2x-3y=3\end{array}\right.\iff \left\{\begin{array}{l}y=10-3x\\ 2x-3(10-3x)=3\end{array}\right.\iff \left\{\begin{array}{l}y=10-3x\\ 11x-30=3\end{array}\right.\iff \left\{\begin{array}{l}y=1\\ x=3\end{array}\right.$

Vậy hệ phương trình có nghiệm duy nhất là (x;y)=(3;1)

$$\begin{gathered} \left\{\begin{array}{l}\left(x+y\right)+\left(x+2y\right)=-2\\ 3\left(x+y\right)+\left(x-2y\right)=1\end{array}\right.\iff \left\{\begin{array}{l}2x+3y=-2\\ 4x+y=1\end{array}\right.\iff \left\{\begin{array}{l}2x+3y=-2\\ -12x-3y=-3\end{array}\iff \left\{\begin{array}{l}2x+3y=-2\\ -10x=-5\end{array}\iff \left\{\begin{array}{l}2.\frac{1}{2}+3y=-2\\ x=\frac{1}{2}\end{array}\right.\right.\right. \\ \iff \left\{\begin{array}{l}x=\frac{1}{2}\\ y=-1\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có nghiệm (x;y)=($\frac{1}{2};-1$)

$\left\{\begin{array}{l}-m-\sqrt{3}=n\\ -n+\sqrt{3}m=1\end{array}\right.\iff \left\{\begin{array}{l}m+n=-\sqrt{3}\\ \sqrt{3}m-n=1\end{array}\right.\iff \left\{\begin{array}{l}m+n=-\sqrt{3}\\ (1+\sqrt{3})m=1-\sqrt{3}\end{array}\right.\iff \left\{\begin{array}{l}m+n=-\sqrt{3}\\ m=\frac{1-\sqrt{3}}{1+\sqrt{3}}\end{array}\right.$

$\iff \left\{\begin{array}{l}m=\sqrt{3}-2\\ n=-\sqrt{3}-m\end{array}\right.\iff \left\{\begin{array}{l}m=\sqrt{3}-2\\ n=-2\sqrt{3}+2\end{array}\right.$

Vậy khi $(m;n)=(\sqrt{3}-2;-2\sqrt{3}+2)$ thì hệ phương trình có nghiệm $(x;y)=(-1;\sqrt{3})$

$\left\{\begin{array}{l}2x=4\\ x+y=5\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=3\end{array}\right.$

Vậy hệ phương trình có nghiệm duy nhất là (x;y)=(2;3)

$\left\{\begin{array}{c}3x-y=1\\ x+2y=5\end{array}\right.\iff \left\{\begin{array}{c}6x-2y=2\\ x+2y=5\end{array}\right.\iff \left\{\begin{array}{c}7x=7\\ 3x-y=1\end{array}\right.\iff \left\{\begin{array}{c}x=1\\ y=3x-1\end{array}\right.\iff \left\{\begin{array}{c}x=1\\ y=2\end{array}\right..$

Vậy hệ phương trình có nghiệm duy nhất là (x;y)=(1;2)

$$\begin{gathered} \left\{\begin{array}{c}2x-y=5\\ x+5y=-3\end{array}\right. \\ \left\{\begin{array}{c}2x-y=5\\ 2x+10y=-6\end{array}\right. \\ \left\{\begin{array}{c}2x-y=5\\ 11y=-11\end{array}\right. \\ \left\{\begin{array}{c}2x=5+y\\ 11y=-11\end{array}\right. \\ \left\{\begin{array}{c}2x=5-1\\ y=-1\end{array}\right. \\ \left\{\begin{array}{c}x=2\\ y=-1\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có một nghiệm duynhaats (x;y)=(2;-1)

$P={\left(x+y\right)}^{2017}={\left[2+\left(-1\right)\right]}^{2017}=1^{2017}=1$

$$\begin{gathered} \left\{\begin{array}{l}3x-2y=5\\ 2x+y=8.\end{array}\right. \\ \iff \left\{\begin{array}{l}3x-2\left(8-2x\right)=5\\ y=8-2x\end{array}\right. \\ \iff \left\{\begin{array}{l}3x-16+4x=5\\ y=8-2x\end{array}\right. \\ \iff \left\{\begin{array}{l}7x=21\\ y=8-2x\end{array}\right. \\ \iff \left\{\begin{array}{l}x=3\\ y=8-2.3\end{array}\right. \\ \iff \left\{\begin{array}{l}x=3\\ y=2\end{array}\right. \end{gathered}$$

Hệ phương trình có nghiệm duy nhất (x;y)=(3;2)

$\begin{array}{l}\left\{\begin{array}{l}5x-2y=11\\ x+y=-2\end{array}\right.\\ \iff \left\{\begin{array}{l}5x-2(-2-x)=11\\ y=-2-x\end{array}\right.\\ \iff \left\{\begin{array}{l}x=1\\ y=-3\end{array}\right.\end{array}$

Vậy hệ phương trình có một nghiệm (1;-3)

Giải hệ phương trình: 

$\left\{\begin{array}{l}3x-2y=6\\ x+2y=10\end{array}\iff \right.\left\{\begin{array}{l}4x=16\\ x+2y=10\end{array}\right.\iff \left\{\begin{array}{l}x=4\\ 4+2y=10\end{array}\right.\iff \left\{\begin{array}{l}x=4\\ y=3\end{array}\right.$

Vậy nghiệm của hệ phương trình là: (x;y)=(4;3)

$$\begin{gathered} \left\{\begin{array}{c}x+y=m\\ x^2+y^2=-m^2+6\end{array}\right. \\ \iff \left\{\begin{array}{c}y=m-x\\ x^2+y^2=-m^2+6\end{array}\right.\iff \left\{\begin{array}{c}y=m-x\\ x^2+{\left(m-x\right)}^2=-m^2+6\end{array}\right.\iff \left\{\begin{array}{c}y=m-x\\ x^2+m^2-2mx+x^2=-m^2+6\end{array}\right. \\ \iff \left\{\begin{array}{c}y=m-x\\ 2x^2-2mx+2m^2-6=0\end{array}\right.\iff \left\{\begin{array}{c}y=m-x\\ x^2-mx+m^2-3=0\end{array}\right. \end{gathered}$$

Hệ phương trình đã cho có nghiệm $\iff$phương trình $x^2-mx+m^2-3=0$ có nghiệm

$$\begin{gathered} \iff \Delta =m^2-4\left(m^2-3\right)\ge 0\iff m^2-4m^2+12\ge 0 \\ \iff 12-3m^2\ge 0\iff m^2\le 4 \\ \iff -2\le m\le 2 \end{gathered}$$

Với m thỏa mãn $-2\le m\le 2$ thì phương trình có nghiệm (x;y). Khi đó ta có:

$$\begin{gathered} P=xy+2\left(x+y\right)=\frac{1}{2}\left[{\left(x+y\right)}^2-\left(x^2+y^2\right)\right]+2\left(x+y\right) \\ \iff P=\frac{1}{2}\left[m^2-\left(-m^2+6\right)\right]+2m=\frac{1}{2}\left(2m^2-6\right)+2m \\ \iff P=m^2+2m-3=m^2+2m+1-4={\left(m+1\right)}^2-4 \end{gathered}$$

Nhận xét: ${\left(m+1\right)}^2\ge 0 \forall m\in \left[-2;2\right]$, dấu bằng xảy ra $\iff m=-1$ thỏa mãn điều kiện

$\Rightarrow P\ge -4$

Dấu bằng xảy ra $\iff m=-1$

Vậy minP=-4 khi m=-1

$\left\{\begin{array}{l}3x-2y=9\\ x-3y=10\end{array}\right.\iff \left\{\begin{array}{l}3x-2y=9\\ 3x-9y=30\end{array}\right.\iff \left\{\begin{array}{l}3x-2y=9\\ 7y=-21\end{array}\right.\iff \left\{\begin{array}{l}3x-2\left(-3\right)=9\\ y=-3\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=-3\end{array}\right.$

Vậy hệ phương trình đã cho có nghiệm duy nhất (1; -3)

$$\begin{gathered} \left\{\begin{array}{c}x+y=m\\ x^2+y^2=-m^2+6\end{array}\right. \\ \iff \left\{\begin{array}{c}y=m-x\\ x^2+y^2=-m^2+6\end{array}\right.\iff \left\{\begin{array}{c}y=m-x\\ x^2+{\left(m-x\right)}^2=-m^2+6\end{array}\right.\iff \left\{\begin{array}{c}y=m-x\\ x^2+m^2-2mx+x^2=-m^2+6\end{array}\right. \\ \iff \left\{\begin{array}{c}y=m-x\\ 2x^2-2mx+2m^2-6=0\end{array}\right.\iff \left\{\begin{array}{c}y=m-x\\ x^2-mx+m^2-3=0\end{array}\right. \end{gathered}$$

Hệ phương trình đã cho có nghiệm $\iff$phương trình $x^2-mx+m^2-3=0$ có nghiệm

$$\begin{gathered} \iff \Delta =m^2-4\left(m^2-3\right)\ge 0\iff m^2-4m^2+12\ge 0 \\ \iff 12-3m^2\ge 0\iff m^2\le 4 \\ \iff -2\le m\le 2 \end{gathered}$$

Với m thỏa mãn $-2\le m\le 2$ thì phương trình có nghiệm (x; y). Khi đó ta có:

$P=xy+2\left(x+y\right)=\frac{1}{2}\left[{\left(x+y\right)}^2-\left(x^2+y^2\right)\right]+2\left(x+y\right)$

$$\begin{gathered} \iff P=\frac{1}{2}\left[m^2-\left(-m^2+6\right)\right]+2m=\frac{1}{2}\left(2m^2-6\right)+2m \\ \iff P=m^2+2m-3=m^2+2m+1-4={\left(m+1\right)}^2-4 \end{gathered}$$

$\left\{\begin{array}{l}{x}^2+2y^2=3\left(1\right)\\ 3x-y^2=2\left(2\right)\end{array}\right.$

$\left\{\begin{array}{l}x+2y=4\\ 3x-y=5\end{array}\right.\iff \left\{\begin{array}{l}x+2y=4\\ 6x-2y=10\end{array}\right.\iff \left\{\begin{array}{l}7x=14\\ 3x-y=5\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$

Vậy hệ có một nghiệm (x;y)=(2;1)

Giải hệ phương trình:

$\left\{\begin{array}{l}2x-y=5\\ x+y=4\end{array}\right.\iff \left\{\begin{array}{l}3x=9\\ x+y=4\end{array}\right.\iff \left\{\begin{array}{l}x=3\\ 3+y=4\end{array}\right.\iff \left\{\begin{array}{l}x=3\\ y=1\end{array}\right.$

Vậy nghiệm của hệ phương trình là: (x;y)=(3;1)

$\left\{\begin{array}{l}7x-3y=4\\ 4x+y=5\end{array}\right.\iff \left\{\begin{array}{l}7x-3y=4\\ 12x+3y=15\end{array}\right.\iff \left\{\begin{array}{l}7x-3y=4\\ 19x=19\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ 7.1-3y=4\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=1\end{array}\right.$

Ta có $\left\{\begin{array}{l}x+2y=3\\ 2x-y=-4\end{array}\right.\iff \left\{\begin{array}{l}x=3-2y\\ 2\left(3-2y\right)-y=-4\end{array}\right.\iff \left\{\begin{array}{l}x=3-2y\\ -5y=-10\end{array}\right.\iff \left\{\begin{array}{l}x=-1\\ y=2\end{array}\right.$

Vậy hệ phương trình đã cho có nghiệm duy nhất là (-1;2)

$\left\{\begin{array}{l}2x+3y=8\\ x+3y=1\end{array}\right.\iff \left\{\begin{array}{l}x=7\\ y=\frac{1-x}{3}=\frac{1-7}{3}=-2\end{array}\right.$

Vậy nghiệm của hệ phương trình là (x;y)=(7;-2)

$\left\{\begin{array}{l}3x+y=5\\ 3-x=y\end{array}\right.\iff \left\{\begin{array}{l}3x+3-x=5\\ y=3-x\end{array}\right.\iff \left\{\begin{array}{l}2x=2\\ y=3-x\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=2\end{array}\right.$

$\left\{\begin{array}{l}3\text{x}-y=7\\ 5\text{x}+y=9\end{array}\right.\iff \left\{\begin{array}{l}3\text{x}-y=7\\ 8\text{x}=16\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=-1\end{array}\right.$

$\left\{\begin{array}{l}3x-2y=10\\ x+3y=7\end{array}\right.\iff \left\{\begin{array}{l}3x-2y=10\\ 3x+9y=21\end{array}\right.\iff \left\{\begin{array}{l}11y=11\\ x+3y=7\end{array}\right.\iff \left\{\begin{array}{l}y=1\\ x+3=7\end{array}\right.\iff \left\{\begin{array}{l}x=4\\ y=1\end{array}\right.$

$\left\{\begin{array}{l}3x-2\left(2y-1\right)=0\\ 3x+2y=2\left(7-x\right)\end{array}\right.\iff \left\{\begin{array}{l}3x-4y=-2\\ 5x+2y=14\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ 2y=14-10\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=2\end{array}\right.$

Vậy hệ phương trình đã cho có nghiệm $\left\{\begin{array}{l}x=2\\ y=2\end{array}\right.$

$\begin{array}{l}\left\{\begin{array}{l}x-2y=3-m\\ 2x+y=3(m+2)\end{array}\right.\iff \left\{\begin{array}{l}2x-4y=6-2m\\ 2x+y=3m+6\end{array}\right.\iff \left\{\begin{array}{l}x-2y=3-m\\ 5y=5m\end{array}\right.\\ \iff \left\{\begin{array}{l}x-2m=3-m\\ y=m\end{array}\right.\iff \left\{\begin{array}{l}x=m+3\\ y=m\end{array}\right.\end{array}$

Do đó:

$\begin{array}{l}A=x^2+y^2=+m^2=2m^2+6m+9\\ =2{\left(m+\frac{3}{2}\right)}^2+\frac{9}{2}\ge \frac{9}{2}\forall m\end{array}$

$\left\{\begin{array}{l}x+3y=9\\ x-y=1\end{array}\right.\iff \left\{\begin{array}{l}4y=8\\ x-y=1\end{array}\right.\iff \left\{\begin{array}{l}y=2\\ x-y=1\end{array}\right.\iff \left\{\begin{array}{l}y=2\\ x=3\end{array}\right.$

Vậy hệ phương trình có nghiệm duy nhất (x;y)=(3;2)

Hệ phương trình đã cho tương đương

$\left\{\begin{array}{l}2x-4y=8\\ 2x+3y=1\end{array}\right.\iff \left\{\begin{array}{l}-7y=7\\ 2x+3y=1\end{array}\right.\iff \left\{\begin{array}{l}y=-1\\ 2x+3\left(-1\right)=1\end{array}\right.\iff \left\{\begin{array}{l}y=-1\\ 2x-3=1\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=-1\end{array}\right.$

Vậy, hệ phương trình có nghiệm duy nhất $\left(x;y\right)=\left(2;-1\right)$.

$\left\{\begin{array}{l}2x-y=4\\ 4x+y=-1\end{array}\right.\iff \left\{\begin{array}{l}6x=3\\ 4x+y=-1\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{1}{2}\\ y=-3\end{array}\right.$

Vậy hệ phương trình có nghiệm $\left(x;y\right)=\left(\frac{1}{2};-3\right)$

$\left\{\begin{array}{l}x+2y=4\\ 5x-2y=8\end{array}\right.\iff \left\{\begin{array}{l}6x=12\\ x+2y=4\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ x+2y=4\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$

Vậy hệ phương trình đã cho có 1 nghiệm (x;y)=(2;1)

$\left\{\begin{array}{l}4x-y=7\\ x+3y=5\end{array}\right.\iff \left\{\begin{array}{l}12\text{x}-3y=21\\ x+3y=5\end{array}\right.\iff \left\{\begin{array}{l}13\text{x}=26\\ y=4\text{x}-7\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=4.\text{2}-7\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$

Vậy hệ phương trình có nghiệm duy nhất là: (x;y)=(2;1)

a)  Ta có

$$\begin{gathered} \left\{\begin{array}{l}2x-y=3\\ x^2+y=5\end{array}\right. \\ \iff \left\{\begin{array}{l}y=2x-3\\ x^2+y=5\end{array}\right. \\ \iff \left\{\begin{array}{l}y=2x-3\\ x^2+2x-3-5=0\end{array}\right. \\ \iff \left\{\begin{array}{l}y=2x-3\\ x^2+2x-8=0\end{array}\right. \\ \iff \left\{\begin{array}{l}y=2x-3\\ \left[\begin{array}{l}x=2\\ x=-4\end{array}\right.\end{array}\right. \\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}x=2\\ y=2.2-3=1\end{array}\right.\\ \left\{\begin{array}{l}x=-4\\ y=2.\left(-4\right)-3=-11\end{array}\right.\end{array}\right. \end{gathered}$$

Vậy hệ có hai nghiệm $\left(x;y\right)\in \left\{\left(2;1\right),\left(-4;-11\right)\right\}$.

$\left\{\begin{array}{l}2x+y=-1\\ x-2y=7\end{array}\right..\iff \left\{\begin{array}{l}4x+2y=-2\\ x-2y=7\end{array}\right.\iff \left\{\begin{array}{l}y=-1-2x\\ 5x=5\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=-3\end{array}\right.$

Vậy hệ đã cho có một nghiệm (1;-3)

$\left\{\begin{array}{l}x-y=1\\ x+y=3\end{array}\right.\iff \left\{\begin{array}{l}2x=4\\ 2y=2\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$

Điều kiện: $a\ne 0,b\ne 0$.

Từ giả thiết, ta có: $\left\{\begin{array}{l}3a^2+\frac{1}{b}=b^2\left(1\right)\\ 3b^2+\frac{2}{a}=a^2\left(2\right)\end{array}\right.$ 

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}3{\text{a}}^{2}b+1=b^3\\ 3b^{2}a+2=a^3\end{array}\right.\iff \left\{\begin{array}{l}3{\text{a}}^{2}b-b^3=-1\\ 3b^{2}a-a^3=-2\end{array}\right.\iff \left\{\begin{array}{l}{\text{9a}}^4{b}^2-6{\text{a}}^2{b}^4+b^6=1\\ 9b^4{a}^2-6b^2{a}^4+a^6=4\end{array}\right.\\ \Rightarrow a^6+3{\text{a}}^4{b}^2+3{\text{a}}^2{b}^4+b^6=5\\ \Rightarrow {\left(a^2+b^2\right)}^3=5\Rightarrow a^2+b^2=\sqrt[3]{5}\end{array}$

$\left\{\begin{array}{l}5x+7y=3\\ 5x-4y=-8\end{array}\right.\iff \left\{\begin{array}{l}11y=11\\ 5x-4y=-8\end{array}\right.\iff \left\{\begin{array}{l}y=1\\ 5x=-4\end{array}\right.\iff \left\{\begin{array}{l}x=-\frac{4}{5}\\ y=1\end{array}\right.$

Vậy hệ phương trình có nghiệm duy nhất $\left(x;y\right)=\left(-\frac{4}{5};1\right)$

$\left\{\begin{array}{l}2x-y=3\\ x+2y=-1\end{array}\right.\iff \left\{\begin{array}{l}4x-2y=6\\ x+2y=-1\end{array}\right.\iff \left\{\begin{array}{l}2x-y=3\\ 5x=5\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=-1\end{array}\right.$

Vậy hệ phương trình có nghiệm duy nhất (x; y)=(1;-1)