a)Ta có: \(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)

\(\Leftrightarrow\dfrac{x}{2}-\dfrac{3x-13}{5}=\dfrac{-7}{5}-\dfrac{7x}{10}\)

\(\Leftrightarrow\dfrac{5x}{10}-\dfrac{2\left(3x-13\right)}{10}=\dfrac{-14}{10}-\dfrac{7x}{10}\)

\(\Leftrightarrow5x-6x+26=-14-7x\)

\(\Leftrightarrow-x+26+14+7x=0\)

\(\Leftrightarrow6x=-40\)

hay \(x=-\dfrac{20}{3}\)

d) Ta có: \(\dfrac{2x-3}{2}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}+\dfrac{-9}{6}=\dfrac{5-3x}{6}-\dfrac{2}{6}\)

\(\Leftrightarrow6x-9-9=5-3x-2\)

\(\Leftrightarrow6x-18-3+3x=0\)

\(\Leftrightarrow x=\dfrac{7}{3}\)

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:

ĐKXĐ: ...

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)

Đặt \(4x+\frac{7}{x}-10=t\)

\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)

\(a,=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\\ b,=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-1\right)}\cdot\dfrac{x-1}{x\left(x+2\right)}=\dfrac{x-2}{x^2}\)

a) \(\left(x^2-25\right):\dfrac{2x+10}{3x-7}=\left(x-5\right)\left(x+5\right).\dfrac{3x-7}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)

b) \(\dfrac{x^2-4}{x^2-x}:\dfrac{x^2+2x}{x-1}=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}.\dfrac{x-1}{x\left(x+2\right)}=\dfrac{x-1}{x^2}\)

a) ĐKXĐ: \(10-5x< 0\Leftrightarrow5x>10\Leftrightarrow x>2\)

b) ĐKXĐ: \(7-3x>0\Leftrightarrow3x< 7\Leftrightarrow x< \dfrac{7}{3}\)

c) ĐKXĐ: \(-5-2x\ge0\Leftrightarrow2x\le-5\Leftrightarrow x\le-\dfrac{5}{2}\)

a) \(x>2\)

b) \(x< \dfrac{7}{3}\)

c) \(x\le-\dfrac{5}{2}\)

a: ĐKXĐ: x>2

b: ĐKXĐ: \(x< \dfrac{7}{3}\)

`x/2-(3x/5-13/5)=7/5-7/10x`

`=>x/2-3x/5+13/5=7/5+(7x)/10`

`=>(5x)/10-(6x)/10-(7x)/10=7/5-13/5`

`=>(5x-6x-7x)/10=(7-13)/5`

`=>(-4x)/5=-6/5`

`=>4x=6`

`=>x=3/2`