\(\dfrac{-13}{15}=\dfrac{-13}{5}\cdot\dfrac{1}{3}=\dfrac{-13}{5}:3\)

\(\dfrac{-2}{11}=-2\cdot\dfrac{1}{11}=-2:11\)

A. không có x

B. 

\(\left(x-73\right)\times10^2-26=8^2+10\\ \left(x-73\right)\times100-26=64+10\\ \left(x-73\right)\times100-26=74\\ \left(x-73\right)\times100=26+74=100\\ x-73=1\\ x=1+73=74\)

C. 

\(\left(x+1\right)^3-2^5:2^3=60\\ \left(x+1\right)^3-2^2=60\\ \left(x+1\right)^3-4=60\\ \left(x+1\right)^3=60+4=64\\ \left(x+1\right)^3=4^3\\ x+1=4\\ x=4-1\\ x=3\)

D. 

\(\left(x+1\right)^4-1=80\\ \left(x+1\right)^4=80+1\\ \left(x+1\right)^4=3^4\)

TH1: 

\(x+1=3\\ x=3-1=2\)

TH2: 

\(x+1=-3\\ x=-3-1=-4\)

A sai đề ạ

a) ĐKXĐ: `x\ne3;x\ne-1`

`\frac{x}{2(x-3)}+\frac{x}{2x+2}=\frac{2x}{(x+1)(x-3)}`

`\Leftrightarrow \frac{x(x+1)}{2(x+1)(x-3)}+\frac{x(x-3)}{2(x+1)(x-3)}=\frac{4x}{2x(x+1)(x-3)}`

`\Rightarrow x(x+1)+x(x-3)=4x`

`\Leftrightarrow 2x^2-2x=4x`

`\Leftrightarrow 2x^2-6x=0`

`\Leftrightarrow 2x(x-3)=0`

\(\Leftrightarrow \left[\begin{array}{} x=0(tm)\\x=3 (ktm)\end{array} \right.\)

b) ĐKXĐ: `x\ne-2`

`\frac{3x}{x^2-2x+4}=\frac{3}{x+2}+\frac{72}{x^3+8}`

`\Leftrightarrow \frac{3x(x+2)}{(x+2)(x^2-2x+4)}=\frac{3(x^2-2x+4)}{(x+2)(x^2-2x+4)}+\frac{72}{(x+2)(x^2-2x+4)}`

`\Rightarrow 3x^2+6x=3x^2-6x+12+72`

`\Leftrightarrow 12x=84`

`\Leftrightarrow x=7(tm)`

$\mathtt{Toru}$

\(a.\dfrac{3}{4}:x=-\dfrac{1}{2}\\ x=\dfrac{3}{4}:\left(-\dfrac{1}{2}\right)\\ x=-\dfrac{3}{2}\\ b.x:\left(-\dfrac{2}{3}\right)=-\dfrac{5}{6}\\ x=\left(-\dfrac{5}{6}\right)\cdot\left(-\dfrac{2}{3}\right)\\ x=\dfrac{5}{9}\\ c.\dfrac{1}{2}:x-\dfrac{1}{2}=-\dfrac{3}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}\\ x=\dfrac{1}{2}:\left(-\dfrac{1}{4}\right)\\ x=-2\\ d.\dfrac{1}{2}\cdot x+\dfrac{3}{4}=-\dfrac{5}{6}\\ \dfrac{1}{2}\cdot x=-\dfrac{19}{12}\\ x=\dfrac{-19}{12}:\dfrac{1}{2}=-\dfrac{19}{12}\cdot2\\ x=-\dfrac{19}{6}\)

\(e.x:\dfrac{1}{2}+\dfrac{1}{2}=-\dfrac{5}{6}\\ x:\dfrac{1}{2}=-\dfrac{4}{3}\\ x=-\dfrac{4}{3}\cdot\dfrac{1}{2}\\ x=-\dfrac{2}{3}\\ g.\dfrac{4}{3}-\dfrac{1}{2}\cdot x=\dfrac{1}{2}\\ \dfrac{1}{2}\cdot x=\dfrac{5}{6}\\ x=\dfrac{5}{6}:\dfrac{1}{2}=\dfrac{5}{6}\cdot2\\ x=\dfrac{5}{3}\\ h.\dfrac{3}{8}\cdot x-\dfrac{1}{2}=\dfrac{3}{4}\\ \dfrac{3}{8}\cdot x=\dfrac{1}{4}\\ x=\dfrac{1}{4}:\dfrac{3}{8}=\dfrac{1}{4}\cdot\dfrac{8}{3}\\ x=\dfrac{2}{3}\)

mình làm lại câu h xíu

\(\dfrac{3}{8}\cdot x-\dfrac{1}{2}=\dfrac{3}{4}\\ \dfrac{3}{8}\cdot x=\dfrac{5}{4}\\ x=\dfrac{5}{4}:\dfrac{3}{8}=\dfrac{5}{4}\cdot\dfrac{8}{3}\\ x=\dfrac{10}{3}\)

a) 

\(\dfrac{4}{x+3}-\dfrac{3}{x-5}=0\left(x\ne-3;x\ne5\right)\\ \Leftrightarrow\dfrac{4\left(x-5\right)}{\left(x+3\right)\left(x-5\right)}-\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x-5\right)}=0\\ \Leftrightarrow4\left(x-5\right)-3\left(x+3\right)=0\\ \Leftrightarrow4x-20-3x-9=0\\ \Leftrightarrow x-29=0\\ \Leftrightarrow x=29\left(tm\right)\)

b) 

\(\dfrac{1}{x+2}-\dfrac{1}{x-2}=\dfrac{3x-12}{x^2-4}\left(x\ne\pm2\right)\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x-2}=\dfrac{3x-12}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x-12}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow x-2-x-2=3x-12\\ \Leftrightarrow-4=3x-12\\ \Leftrightarrow3x=-4+12\\ \Leftrightarrow3x=8\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)

c) 

\(\dfrac{1}{x+1}-\dfrac{4}{x^2-x+1}=\dfrac{2x^2+1}{x^3+1}\left(x\ne-1\right)\\ \Leftrightarrow\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}\\ \Leftrightarrow x^2-x+1-4x-4=2x^2+1\\ \Leftrightarrow x^2-5x-3=2x^2+1\\ \Leftrightarrow2x^2-x^2+5x+1+3=0\\ \Leftrightarrow x^2+5x+4=0\\ \Leftrightarrow\left(x+1\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

a) Đặt: `4x-5=t`

\(t^2+2t+1=0\\ \Leftrightarrow\left(t+1\right)^2=0\\ \Leftrightarrow t+1=0\\ \Leftrightarrow t=-1\)

\(\Rightarrow4x-5=-1\\ \Leftrightarrow4x=-1+5\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\) 

b) Đặt: \(x^2-x=t\)

\(t\left(t+1\right)=6\\ \Leftrightarrow t^2+t-6=0\\ \Leftrightarrow t^2-2t+3t-6=0\\ \Leftrightarrow t\left(t-2\right)+3\left(t-2\right)=0\\ \Leftrightarrow\left(t-2\right)\left(t+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\)

Với: 

\(t=2\Rightarrow x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Với: 

\(t=-3\Rightarrow x^2-x=-3\Leftrightarrow x^2-x+3=0\)

Mà: `x^2-x+3>0` nên vô lý 

\(a.\dfrac{3}{2}-4\cdot\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\\ =\dfrac{3}{2}-4\cdot\dfrac{5}{4}\\ =\dfrac{3}{2}-5=-\dfrac{7}{2}\\ b.\left(-\dfrac{1}{3}+\dfrac{5}{6}\right)\cdot11-7\\ =\dfrac{1}{2}\cdot11-7\\ =\dfrac{11}{2}-7=-\dfrac{3}{2}\\ c.\left(\dfrac{5}{8}+\dfrac{-2}{7}\right):\dfrac{3}{4}+\left(\dfrac{3}{8}-\dfrac{5}{7}\right):\dfrac{3}{4}\\ =\left(\dfrac{5}{8}+\dfrac{-2}{7}\right)\cdot\dfrac{4}{3}+\left(\dfrac{3}{8}-\dfrac{5}{7}\right)\cdot\dfrac{4}{3}\\ =\dfrac{4}{3}\cdot\left(\dfrac{5}{8}-\dfrac{2}{7}+\dfrac{3}{8}-\dfrac{5}{7}\right)\)

\(\dfrac{4}{3}\cdot\left[\left(\dfrac{5}{8}+\dfrac{3}{8}\right)-\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\right]\\ =\dfrac{4}{3}\cdot\left(1-1\right)=\dfrac{4}{3}\cdot0=0\)

\(d.\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\\ =\dfrac{3\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

a) 

\(\dfrac{3}{2}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\\ =\dfrac{3}{4}-4\left(\dfrac{2}{4}+\dfrac{3}{4}\right)\\ =\dfrac{3}{4}-4\cdot\dfrac{5}{4}\\ =\dfrac{3}{4}-5\\ =-\dfrac{17}{4}\) 

b) 

\(\left(-\dfrac{1}{3}+\dfrac{5}{6}\right)\times11-7\\ =\left(-\dfrac{2}{6}+\dfrac{5}{6}\right)\times11-7\\ =\dfrac{1}{2}\times11-7\\ =\dfrac{11}{2}-7\\ =-\dfrac{3}{2}\)

c) 

\(\left(\dfrac{5}{8}+\dfrac{-2}{7}\right):\dfrac{3}{4}+\left(\dfrac{3}{8}-\dfrac{5}{7}\right):\dfrac{3}{4}\\ =\left(\dfrac{5}{8}+\dfrac{-2}{7}\right)\times\dfrac{4}{3}+\left(\dfrac{3}{8}-\dfrac{5}{7}\right)\times\dfrac{4}{3}\\ =\dfrac{4}{3}\times\left(\dfrac{5}{8}+\dfrac{-2}{7}+\dfrac{3}{8}-\dfrac{5}{7}\right)\times\dfrac{4}{3}\\ =\left(1-1\right)\times\dfrac{4}{3}\\ =0\times\dfrac{4}{3}\\ =0\)

d) 

\(M=\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\\ =\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}\\ =\dfrac{3}{13}\)

a) 

\(3x\left(x-4\right)+7\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(3x+7\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\3x=-7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{3}\end{matrix}\right.\)

Vậy: ...

b) 

\(\left(4x^2-9\right)+\left(x+2\right)\left(3-2x\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)-\left(x+2\right)\left(2x-3\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3-x-2\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)

Vậy: ...

c) 

\(\left(x-1\right)^2=\left(2x+3\right)^2\\ \Leftrightarrow\left(x-1\right)^2-\left(2x+3\right)^2=0\\ \Leftrightarrow\left(x-1+2x+3\right)\left(x-1-2x-3\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(-x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x=-2\\-x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-4\end{matrix}\right.\)

Vậy: ...

d) 

\(x^2-6x+5=0\\ \Leftrightarrow x^2-5x-x+5=0\\ \Leftrightarrow x\left(x-5\right)-\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

Vậy: ... 

\(a.\left(-\dfrac{9}{34}\right)\cdot\dfrac{17}{4}=-\dfrac{9}{8}\\ b.\left(-\dfrac{20}{41}\right)\cdot\left(-\dfrac{4}{5}\right)=\dfrac{16}{41}\\ c.1\dfrac{1}{7}\cdot\left(-3\dfrac{1}{2}\right)=\dfrac{8}{7}\cdot\left(-\dfrac{7}{2}\right)=-4\\ d.\left(-\dfrac{5}{2}\right):\dfrac{3}{4}=\left(-\dfrac{5}{2}\right)\cdot\dfrac{4}{3}=-\dfrac{10}{3}\\ e.\left(-\dfrac{3}{4}\right):\left(-1\dfrac{1}{3}\right)=\left(-\dfrac{3}{4}\right)\cdot\left(-\dfrac{3}{4}\right)=\dfrac{9}{16}\\ g.\left(-5\right):\left(-2\dfrac{1}{2}\right)=\left(-5\right)\cdot\left(-\dfrac{2}{5}\right)=2\)

\(a)-\dfrac{9}{34}\times\dfrac{17}{4}\\ =\dfrac{-9}{17\times2}\times\dfrac{17}{4}\\ =\dfrac{-9}{8}\\ b)-\dfrac{20}{41}\times-\dfrac{4}{5}\\ =\dfrac{4\times5}{41}\times\dfrac{-4}{5}\\ =-\dfrac{16}{41}\\ c)1\dfrac{1}{7}\times\left(-3\dfrac{1}{2}\right)\\ =\dfrac{8}{7}\times\dfrac{-7}{2}\\ =-4\\ d)-\dfrac{5}{2}:\dfrac{3}{4}\\ =-\dfrac{5}{2}\times\dfrac{4}{3}\\ =\dfrac{-10}{3}\\ e)-\dfrac{3}{4}:\left(-1\dfrac{1}{3}\right)\\ =-\dfrac{3}{4}:-\dfrac{4}{3}\\ =-\dfrac{3}{4}\times-\dfrac{3}{4}\\ =\dfrac{9}{16}\\ g)-5:\left(-2\dfrac{1}{2}\right)\\ =-5:-\dfrac{5}{2}\\ =-5\times-\dfrac{2}{5}\\ =2\)