Dạng 2: Phiếu 1 có đáp án

$\frac{2}{x-1}+\frac{4}{x+3}=\frac{3x+11}{x^2+2x-3}$ (ĐKXĐ: $x\ne 1,x\ne -3$)

$\begin{array}{l}\iff \frac{2\left(x+3\right)+4\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{3x+11}{\left(x-1\right)\left(x+3\right)}\\ \iff \frac{2x+6+4x-4}{\left(x-1\right)\left(x+3\right)}=\frac{3x+11}{\left(x-1\right)\left(x+3\right)}\\ \Rightarrow 6x+2=3x+11\\ \iff 3x=9\\ \iff x=3\left(TM\right)\end{array}$

$S=\left\{3\right\}$

Đặt $2x^2+3x=A$ ta có:

$\begin{array}{l}{\left(A-1\right)}^2-5.\left(A+3\right)+24=0\\ \iff A^2-7A+10=0\\ \iff \left(A-5\right)\left(A-2\right)=0\\ \Rightarrow \left[\begin{array}{l}A-5=0\\ A-2=0\end{array}\right.\iff \left[\begin{array}{l}A=5\\ A=2\end{array}\right.\end{array}$

TH1: Với $A=5\Rightarrow 2x^2+3x=5$

$\begin{array}{l}\iff 2x^2+3x-5=0\\ \iff \left(x-1\right)\left(2x+5\right)=0\\ \iff \left[\begin{array}{l}x=1\\ x=\frac{-5}{2}\end{array}\right.\end{array}$

TH2: Với $A=2\Rightarrow 2x^2+3x=2$

$\begin{array}{l}\iff 2x^2+3x-2=0\\ \iff \left(2x-1\right)\left(x+2\right)=0\\ \Rightarrow \left[\begin{array}{l}x=\frac{1}{2}\\ x=-2\end{array}\right.\end{array}$

$S=\left\{\frac{-5}{2};-2;\frac{1}{2};1\right\}$

$x^3-3x^2+4=0$

$$\begin{gathered} \begin{array}{l}\iff \left(x+1\right)\left(x^2-4x+4\right)=0\\ \iff \left(x+1\right){\left(x-2\right)}^2=0\\ \iff \left[\begin{array}{l}x+1=0\\ {\left(x-2\right)}^2=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=-1\\ x=2\end{array}\right.\end{array} \\ S=\left\{-1;2\right\} \end{gathered}$$

$x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24$

$\begin{array}{l}\iff x.\left(x+3\right).\left(x+1\right).\left(x+2\right)=24\\ \iff \left(x^2+3x\right).\left(x^2+3x+2\right)=24\end{array}$

Đặt $x^2+3x=A$, ta có $A.\left(A+2\right)=24$

$\begin{array}{l}{A}^2+2A-24=0\\ \iff \left(A-4\right)\left(A+6\right)=0\\ \iff \left[\begin{array}{l}A-4=0\\ A+6=0\end{array}\right.\iff \left[\begin{array}{l}A=4\\ A=-6\end{array}\right.\end{array}$

TH1: $A=4\Rightarrow x^2+3x-4=0$

$\begin{array}{l}\iff \left(x-1\right)\left(x+4\right)=0\\ \iff \left[\begin{array}{l}x-1=0\\ x+4=0\end{array}\right.\iff \left[\begin{array}{l}x=1\\ x=-4\end{array}\right.\end{array}$

TH2: $A=-6\Rightarrow x^2+3x+6=0$

Ta có: 

$x^2+3x+6=x^2+2.\frac{3}{2}.x+\frac{9}{4}-\frac{9}{4}+6={\left(x+\frac{3}{2}\right)}^2+\frac{15}{4}>0\forall x\in ℝ$

$\frac{x+15}{60}+\frac{x+14}{61}=\frac{x+13}{62}+\frac{x+12}{63}$

$\begin{array}{l}\iff \frac{x+15}{60}+1+\frac{x+14}{61}+1=\frac{x+13}{62}+1+\frac{x+12}{63}+1\\ \iff \frac{x+75}{60}+\frac{x+75}{61}=\frac{x+75}{62}+\frac{x+75}{63}\\ \iff \left(x+75\right).\left(\frac{1}{60}+\frac{1}{61}-\frac{1}{62}-\frac{1}{63}\right)=0\end{array}$

Vì $\frac{1}{60}+\frac{1}{61}-\frac{1}{62}-\frac{1}{63}\ne 0\Rightarrow x+75=0\Rightarrow x=-75$

$S=\left\{-75\right\}$

$\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}$ (ĐKXĐ: $x\ne -9;x\ne -10$)

$$\begin{gathered} \begin{array}{l}\iff \frac{x+9}{10}+1+\frac{x+10}{9}+1=\frac{9}{x+10}+1+\frac{10}{x+9}+1\\ \iff \frac{x+19}{10}+\frac{x+19}{9}=\frac{x+19}{x+10}+\frac{x+19}{x+9}\\ \iff \left(x+19\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}\right)=0\\ \iff \left[\begin{array}{l}x+19=0\\ \frac{1}{10}+\frac{1}{9}-\frac{1}{x+10}-\frac{1}{x+9}=0\end{array}\right.\iff \left[\begin{array}{l}x=-19\left(TM\right)\\ x=0\left(TM\right)\end{array}\right.\end{array} \\ S=\left\{-19;0\right\} \end{gathered}$$

$\begin{array}{l}dk:2-x\ge 0\iff x\le 2\\ PT\iff \left[\begin{array}{l}2x-5=2-x\\ 2x-5=-\left(2-x\right)\end{array}\right.\end{array}$$\iff \left[\begin{array}{l}3x=7\\ x=3\end{array}\right.\iff \left[\begin{array}{l}x=\frac{7}{3}\\ x=3\end{array}\right.\left(loại\right)$

$S=\varnothing$

$\left|3x-2\right|+x=11$$\iff \left|3x-2\right|=11-x$

$$\begin{gathered} \iff \begin{array}{l}DK:11-x\ge 0\iff x\le 11\\ PT\iff \left[\begin{array}{l}3x-2=11-x\\ 3x-2=-\left(11-x\right)\end{array}\right.\end{array}\iff \left[\begin{array}{l}4x=13\\ 2x=-9\end{array}\right. \\ \iff \left[\begin{array}{l}x=\frac{13}{4}\\ x=\frac{-9}{2}\end{array}\right.\left(tm\right) \end{gathered}$$

$S=\left\{\frac{-9}{2};\frac{13}{4}\right\}$

$\left|2x-1\right|=1-2x$$\iff \left|2x-1\right|=-(1-2x)\Rightarrow 2x-1\le 0\iff x\le \frac{1}{2}$

Vậy: $S=\left\{x|x\le \frac{1}{2}\right\}$

$\left|-4x\right|=x^2+4$

$\begin{array}{l}DK:x^2+4>0\forall x\in ℝ\\ PT\iff \left[\begin{array}{l}4x=x^2+4\\ 4x=-\left(x^2+4\right)\end{array}\right.\iff \left[\begin{array}{l}{x}^2+4x+4=0\\ x^2-4x+4=0\end{array}\right.\iff \left[\begin{array}{l}{\left(x+2\right)}^2=0\\ {\left(x-2\right)}^2=0\end{array}\right.\iff \left[\begin{array}{l}x=-2\\ x=2\end{array}\right.\end{array}$

$S=\left\{-2;2\right\}$

$\left(2x+1\right).\left(3x-2\right)\le 0$

TH1: $\left\{\begin{array}{l}2x+1\ge 0\\ 3x-2\le 0\end{array}\right.\iff \left\{\begin{array}{l}2x\ge -1\\ 3x\le 2\end{array}\right.\iff \left\{\begin{array}{l}x\ge -\frac{1}{2}\\ x\le \frac{2}{3}\end{array}\right.\iff -\frac{1}{2}\le x\le \frac{2}{3}$

TH2: $\left\{\begin{array}{l}2x+1<0\\ 3x-2>0\end{array}\right.\iff \left\{\begin{array}{l}2x<-1\\ 3x>2\end{array}\right.\iff \left\{\begin{array}{l}x<-\frac{1}{2}\\ x>\frac{2}{3}\end{array}\right.\iff x\in \varnothing$

Vậy: $S=\left\{x\left|-\frac{1}{2}\le x\le \frac{2}{3}\right.\right\}$

Gọi vân tốc riêng của ca nô là $x\left(km/h;x>2\right)$

Vận tốc của ca nô khi đi xuôi dòng là: $x+2\left(km/h\right)$

Vận tốc của ca nô khi đi ngược dòng là $x-2\left(km/h\right)$

Quãng đường ca nô xuôi dòng là $5\left(x+2\right)\left(km\right)$

Quãng đường ca nô ngược dòng là $6\left(x-2\right)\left(km\right)$

Theo đầu bài ta có phương trình:

$\begin{array}{l}5\left(x+2\right)=6\left(x-2\right)\\ \iff 5x+10=6x-12\\ \iff x=22\begin{array}{cc}& \end{array}\left(TM\right)\end{array}$

Vậy quãng đường AB dài là $5.\left(22+2\right)=120km$

Nửa chu vi hình chữ nhật là $\frac{300}{2}=150\left(cm\right)$ 

Gọi chiều dài hình chữ nhật là  x ($150>x>75$ , cm)

Chiều rộng hình chữ nhật là $150-x$ (cm)

Diện tích hình chữ nhật ban đầu là $\text{x}.\left(150-x\right)\left(c{m}^2\right)$ 

Chiều dài sau khi giảm 5cm là : $x-5\left(cm\right)$ 

Chiều rộng sau khi tăng 5cm là $150-x+5=155-x\left(cm\right)$ 

Diện tích hình chữ nhật mới là $\left(x-5\right)\left(155-x\right)\left(c{m}^2\right)$ 

Vì diện tích tăng 275cm2 nên ta có phương trình

$\begin{array}{l}\left(x-5\right)\left(155-x\right)-x(150-x)=275\\ \iff -x^2+160x-755-150x+x^2=275\\ \iff 10x=1030\\ \iff x=103\left(TM\right)\end{array}$

Vậy chiều dài hình chữ nhật là 103 (cm)

Chiều rộng hình chữ nhật là $150-103=47\left(cm\right)$