Bài tập các hằng đẳng thức đáng nhớ

Ta có:

$3599=3600-1={60}^2-1^2=\left(60+1\right)\left(60-1\right)=61.59$

$\begin{array}{l}{x}^2+2{\left(x+1\right)}^2+3{\left(x+2\right)}^2+4{\left(x+3\right)}^2\\ =x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\\ =x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\\ =10x^2+40x+50\\ =\left(x^2+10x+25\right)+\left(9x^2+30x+25\right)\\ ={\left(x+5\right)}^2+{\left(3x+5\right)}^2\end{array}$

Ta có: ${\left(x+y+z\right)}^2=x^2+y^2+z^2+2xy+2yz+2xz$

Mà $x+y+z=0$ và $xy+yz+zx=0$ nên ${}^2+y^2+z^2=0$

$\Rightarrow \left\{\begin{array}{l}{x}^2=0\\ y^2=0\\ z^2=0\end{array}\right.\Rightarrow x=y=z=0$

$$\begin{gathered} A=-1^2+2^2-3^2+4^2-\mathrm{...}-{99}^2+{100}^2 \\ \begin{array}{l}=\left(2^2-1^2\right)+\left(4^2-3^2\right)+\mathrm{...}+\left({100}^2-{99}^2\right)\\ =1+2+3+4+\mathrm{...}+99+100\\ =\frac{\mathrm{100...101}}{2}=5050\end{array} \end{gathered}$$

Xét hai trường hợp

* Nếu n là chẵn thì

$\begin{array}{l}A=\left(2^2-1^2\right)+\left(4^2-3^2\right)+\mathrm{...}+\left({100}^2-{99}^2\right)\\ =1+2+3+4+\mathrm{...}+\left(n-1\right)+n\\ =\frac{n\left(n+1\right)}{2}\end{array}$

*Nếu n là lẻ thì

$\begin{array}{l}A=-1^2+2^2-3^2+4^2-\mathrm{...}+{\left(-1\right)}^n.n^2\\ =\left(2^2-1^2\right)+\left(4^2-3^2\right)+\mathrm{...}+\left[{\left(n-1\right)}^2-{\left(n-2\right)}^2\right]-n^2\\ =1+2+3+4+\mathrm{...}+\left(n-1\right)-n^2\\ =\frac{n\left(n-1\right)}{2}-n^2\\ =-\frac{n\left(n+1\right)}{2}\end{array}$

$\frac{{63}^2-{47}^2}{{215}^2-{105}^2}=\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}=\frac{16.110}{110.320}=\frac{1}{20}$

$\frac{{437}^2-{363}^2}{{537}^2-{463}^2}=\frac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}=\frac{74.800}{74.1000}=\frac{4}{5}$

$A=\left(26-24\right)\left(26+24\right)$ và $B=\left(27-25\right)\left(27+25\right)=\left(26-24\right)\left(26+24+2\right)>A$

$4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-11=0$

Rút gọn ta được $4x+2=0\iff x=-\frac{1}{2}$

$$\begin{gathered} {\left(a+b-c\right)}^2+{\left(a-b+c\right)}^2-2{\left(b-c\right)}^2 \\ =\left(a^2+b^2+c^2+2ab-2ac-2bc\right)+\left(a^2+b^2+c^2-2ab+2ac-2bc\right)+2\left(b^2+c^2-2bc\right) \\ =2a^2 \end{gathered}$$

${\left(a+b+c\right)}^2+{\left(a-b-c\right)}^2+{\left(b-c-a\right)}^2+{\left(c-a-b\right)}^2$

$=\left(a^2+b^2+c^2+2ab+2ac+2bc\right)+\left(a^2+b^2+c^2-2ab-2ac+2bc\right)+\left(a^2+b^2+c^2-2ab+2ac-2bc\right)$

$=4\left(a^2+b^2+c^2\right)$

${\left(a+b+c+d\right)}^2+{\left(a+b-c-d\right)}^2+{\left(a+c-b-d\right)}^2+{\left(a+d-b-c\right)}^2$

$={\left[\left(a+b\right)+\left(c+d\right)\right]}^2+{\left[\left(a+b\right)-\left(c+d\right)\right]}^2+{\left[\left(a+c\right)-\left(b+d\right)\right]}^2+{\left[\left(a+d\right)\left(b+c\right)\right]}^2$

$=2{\left(a+b\right)}^2+2{\left(c+d\right)}^2+{\left(a+c\right)}^2+{\left(b+d\right)}^2+{\left(a+d\right)}^2+{\left(b+c\right)}^2-2\left(ad+bc+ac+bd\right)$

$=4\left(a^2+b^2+c^2+d^2\right)$

Ta có

$\begin{array}{l}A=x^2+2xy+y^2-4x-4y+1\\ ={\left(x+y\right)}^2-4\left(x+y\right)+1\end{array}$

Thay $x+y=3$ vào biểu thức A ta được $A=3^2-4.3+1=-2$

Đặt: $x=a+b+c$ thì 

$A={\left(2x-3c\right)}^2+{\left(2x-3b\right)}^2+{\left(2x-3a\right)}^2$

$\begin{array}{l}=\left(4x^2-12xc+9c^2\right)+\left(4x^2-12xb+9b^2\right)+\left(4x^2-12xa+9a^2\right)\\ =12x^2-12\left(a+b+c\right)+9\left(a^2+b^2+c^2\right)\\ =12x^2-12x^2+9m\\ =9m\end{array}$

a) $899=900-1={30}^2-1^2=\left(30-1\right)\left(30+1\right)=29.31$

b) $9991=10000-9={100}^2-3^2=\left(100-3\right)\left(100+3\right)=97.103$

Ta có

${7778}^2-{2223}^2=\left(7778-2223\right)\left(7778+2223\right)$$=5555.10001=55555555$

$\begin{array}{l}VT={\left(a+b+c\right)}^2+a^2+b^2+c^2\\ =a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\\ =\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)\\ ={\left(a+b\right)}^2+{\left(b+c\right)}^2+{\left(a+c\right)}^2\\ =VP\end{array}$

$\Rightarrow VT=VP$

$\begin{array}{l}VT=x^4+y^4+{\left(x+y\right)}^4\\ =x^4+y^4+{\left(x^2+y^2+2xy\right)}^2\\ =2x^2{y}^2+\left[{\left(x^2+y^2\right)}^2+4xy\left(x^2+y^2\right)+4x^2{y}^2\right]\\ =x^4+y^4+\left(x^4+y^4+4x^2{y}^3+4x^3{y}^2+6x^2{y}^2\right)\\ =2x^4+2y^4+4x^2{y}^3+4x^3{y}^2+4x^2{y}^2+6x^2{y}^2\\ =2\left(x^4+y^4+2x^2{y}^3+2x^3{y}^2+3x^2{y}^2\right)\\ =2\left[\left(x^4+y^4+2x^2{y}^2\right)+2xy\left(x^2+y^2\right)+x^2{y}^2\right]\\ =2{\left(x^2+y^2+xy\right)}^2\\ =VP\\ \Rightarrow VT=VP\end{array}$

Xét hai trường hợp

* Nếu n là chẵn thì

$\begin{array}{l}A=\left(2^2-1^2\right)+\left(4^2-3^2\right)+\mathrm{...}+\left({100}^2-{99}^2\right)\\ =1+2+3+4+\mathrm{...}+\left(n-1\right)+n\\ =\frac{n\left(n+1\right)}{2}\end{array}$

*Nếu n là lẻ thì

$\begin{array}{l}A=-1^2+2^2-3^2+4^2-\mathrm{...}+{\left(-1\right)}^n.n^2\\ =\left(2^2-1^2\right)+\left(4^2-3^2\right)+\mathrm{...}+\left[{\left(n-1\right)}^2-{\left(n-2\right)}^2\right]-n^2\\ =1+2+3+4+\mathrm{...}+\left(n-1\right)-n^2\\ =\frac{n\left(n-1\right)}{2}-n^2\\ =-\frac{n\left(n+1\right)}{2}\end{array}$

$$\begin{gathered} {\left(3a-5b\right)}^2=9a^2+25b^2-30ab \\ \begin{array}{l}=25a^2+9b^2-30ab-16\left(a^2-b^2\right)\\ ={\left(5a\right)}^2+{\left(3b\right)}^2-2.\left(5a\right)\left(3b\right)-16.4c^2\\ ={\left(5a-3b\right)}^2-{\left(8c\right)}^2\\ =\left(5a-3b-8c\right)\left(5a-3b+8c\right)\end{array} \end{gathered}$$

$\left(a^2+b^2\right)\left(x^2+y^2\right)={\left(ax+by\right)}^2$

$\begin{array}{l}\left(a^2+b^2\right)\left(x^2+y^2\right)={\left(ax+by\right)}^2\\ a^2{x}^2+b^2{y}^2+a^2{y}^2+b^2{x}^2=a^2{x}^2+b^2{y}^2+2abxy\\ a^2{y}^2-2abxy+b^2{x}^2=0\\ {\left(ay-bx\right)}^2=0\\ ay-bx=0\\ \frac{a}{x}=\frac{b}{y}\end{array}$

Với $x, y \ne 0$

$$\begin{gathered} \left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)={\left(ax+by+cz\right)}^2 \\ \iff \left(a^2{y}^2-2abxy+b^2{x}^2\right)+\left(a^2{z}^2-2acxz+c^2{x}^2\right)+\left(b^2{z}^2-2bcyz+c^2{y}^2\right)=0 \\ \iff {\left(ay-bx\right)}^2+{\left(az-cx\right)}^2+{\left(bz-cy\right)}^2=0 \\ \iff ay-bx=0,az-cx=0,bz-cy=0 \\ \iff \frac{a}{x}=\frac{b}{y}=\frac{c}{z} \end{gathered}$$

Với $x,y,z\ne 0$

$\begin{array}{l}{\left(a+b\right)}^2=2\left(a^2+b^2\right)\\ a^2+2ab+b^2=2a^2+2b^2\\ 0=a^2-2ab+b^2\\ 0={\left(a-b\right)}^2\\ 0=a-b\\ a=b\end{array}$

$\begin{array}{l}{a}^2+b^2+c^2=ab+bc+ca\\ 2a^2+2b^2+2c^2=2ab+2bc+2ca\\ \left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0\\ \Rightarrow \left\{\begin{array}{l}{\left(a-b\right)}^2=0\\ {\left(b-c\right)}^2=0\\ {\left(c-a\right)}^2=0\end{array}\right.\Rightarrow a=b=c\end{array}$

$\begin{array}{l}{\left(a+b+c\right)}^2=3\left(a^2+b^2+c^2\right)\\ a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\\ 2a^2+2b^2+2c^2=2ab+2bc+2ca\\ \left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0\\ \Rightarrow \left\{\begin{array}{l}{\left(a-b\right)}^2=0\\ {\left(b-c\right)}^2=0\\ {\left(c-a\right)}^2=0\end{array}\right.\Rightarrow a=b=c\end{array}$

${\left(a+b+c\right)}^2=3\left(a^2+b^2+c^2\right)=3\left(ab+bc+ca\right)$

Suy ra $a^2+b^2+c^2=ab+bc+ca$

Suy ra $a=b=c$

$\begin{array}{l}{\left(a+b+c\right)}^2+a^2+b^2+c^2\\ =\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2\right)\\ =\left(a^2+2ab+b^2\right)+\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)\\ ={\left(a+b\right)}^2+{\left(a+b\right)}^2+{\left(b+c\right)}^2\end{array}$

Đặt $x=a-b,y=b-c,x=c-a$ thì biểu thức trở thành $-2xy-2xz-2yz=x^2+y^2+z^2-{\left(x+y+z\right)}^2$$={\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2$

Theo công thức $a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)}{2}$ ta có: $a^4+b^4+c^4=\frac{2^2}{2}=2$

Theo công thức $a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)}{2}$ ta có: $a^4+b^4+c^4=\frac{1^2}{2}=\frac{1}{2}$

1. Bình phương hai vế của a+b+c=0, được:

$a^2+b^2+c^2+2\left(ab+bc+ca\right)=0$

$\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\left(1\right)$

Bình phương hai vế của (1) ta được: 

$a^4+b^4+c^4+2\left(a^2{b}^2+b^2{c}^2+c^2{a}^2\right)$

$=4\left[a^2{b}^2+b^2{c}^2+c^2{a}^2+2abc\left(a+b+c\right)\right]$

$=4\left(a^2{b}^2+b^2{c}^2+c^2{a}^2\right)$

Suy ra $a^4+b^4+c^4=2\left(a^2{b}^2+b^2{c}^2+c^2{a}^2\right)$

2. Bình phương hai vế của (1) ta được

$a^4+b^4+c^4+2\left(a^2{b}^2+b^2{c}^2+c^2{a}^2\right)=4{\left(ab+bc+ca\right)}^2\left(2\right)$

Từ (2) suy ra $2{\left(ab+bc+ca\right)}^2=\frac{a^4+b^4+c^4+2\left(a^2{b}^2+b^2{c}^2+c^2{a}^2\right)}{2}\left(3\right)$

Từ (3) và câu 1 suy ra $a^4+b^4+c^4=2{\left(ab+bc+ca\right)}^2$

3. Bình phương hai vế của (1), chia cho 2,được: $\frac{{\left(a^2+b^2+c^2\right)}^2}{2}=2{\left(ab+bc+ca\right)}^2=a^4+b^4+c^4$

$9x^2-6x+2={\left(3x\right)}^2-2.3x+1+1={\left(3x-1\right)}^2+1>0$

$x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}={\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}>0$

$2x^2+2x+1=x^2+\left(x^2+2x+1\right)=x^2+{\left(x+1\right)}^2>0$

a) $A=x^2-3x+5$

$=x^2-2.\frac{3}{2}x+\text{}\frac{9}{4}+\frac{11}{4}$

$={\left(x-\frac{3}{2}\right)}^2+\frac{11}{4}\ge \frac{11}{4}$

Giá trị nhỏ nhất là $A=\frac{11}{4}$ khi $x=\frac{3}{2}$

b) $B={\left(2x-1\right)}^2+{\left(x+2\right)}^2$

$=5x^2+5\ge 5$

Giá trị nhỏ nhất là B = 5 khi x = 0.

a) $A=4-x^2+2x$

$=5-\left(x^2-2x+1\right)$

$=5-{\left(x-1\right)}^2\le 5$

Vậy GTLN của A là 5

b)

$\begin{array}{l}B=4x-x^2\\ =4-\left(x^2-2.2x+2^2\right)\\ =4-{\left(x-2\right)}^2\le 4\end{array}$

Giá trị lớn nhất của biểu thức là B = 4 khi x = 2.

a) Ta có $999991=1000000-9={1000}^2-3^2=1003.997$ nên là hợp số

b) Ta có $1000027=\left({100}^3+3^3\right)⋮\left(100+3\right)$ nên là hợp số.

$$\begin{gathered} {\left(x-2\right)}^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right) \\ =x^3-6x^2+12x-8-x^3+x+6x^2-18x \\ =-5x-8 \end{gathered}$$ 

$$\begin{gathered} \left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right) \\ =\left(x+2\right)\left(x^2-2x+4\right).\left(x-2\right)\left(x^2+2x+4\right) \\ =\left(x^3+8\right)\left(x^3-8\right) \\ =x^6-64 \end{gathered}$$

$$\begin{gathered} \left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1 \\ {x}^3-3^3+x\left(4-x^2\right)=1 \\ x=7 \end{gathered}$$

$$\begin{gathered} {\left(x+1\right)}^3-{\left(x-1\right)}^3-6{\left(x-1\right)}^2=-10 \\ \iff 6x^2+2-6\left(x^2-2x+1\right)=-10 \\ \iff x=-\frac{1}{2} \end{gathered}$$

$$\begin{gathered} {\left(a+b+c\right)}^3-{\left(b+c-a\right)}^3-{\left(a+c-b\right)}^3-{\left(a+b-c\right)}^3 \\ \begin{array}{l}={\left[a+\left(b+c\right)\right]}^3-{\left[\left(b+c\right)-a\right]}^3-{\left[a-\left(b-c\right)\right]}^3-{\left[a-\left(b-c\right)\right]}^3\\ =6{\left(b+c\right)}^{2}a+2a^3-2a^3-6a{\left(b-c\right)}^2\\ =24abc\end{array} \end{gathered}$$

$$\begin{gathered} {\left(a+b\right)}^3+{\left(b+c\right)}^3+{\left(c+a\right)}^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right) \\ =2a^3+2b^3+2c^3+3\left(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2\right)-3\left(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2+2abc\right) \\ =2\left(a^3+b^3+c^3-3abc\right) \end{gathered}$$

$$\begin{gathered} {\left(a+b+c\right)}^3-a^3-b^3-c^3=\left[{\left(a+b+c\right)}^3-a^3\right]-\left[b^3+c^3\right] \\ \begin{array}{l}=\left(b+c\right)\left[{\left(a+b+c\right)}^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\\ =\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\\ =3\left(a+b\right)\left(b+c\right)\left(c+a\right)\end{array} \end{gathered}$$

$\begin{array}{l}{a}^3+b^3+c^3-3abc\\ ={\left(a+b\right)}^3+c^3-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left[{\left(a+b\right)}^2-c\left(a+b\right)+c^2-3ab\right]\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\end{array}$

Từ giả thiết $\text{a}+b+c=0\Rightarrow c=-\left(a+b\right)$, thay vào đẳng thức cần chứng minh ta được 

${\text{a}}^3+b^3-{\left(a+b\right)}^3=-3ab\left(a+b\right)$$\iff -3a{b}^2-3a^{2}b=-3a{b}^2-3a^{2}b$