a) đk: x khác 1; \(\dfrac{3}{2}\)

 \(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)

= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)

= \(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\dfrac{1-x}{-3x+5}=\dfrac{-1}{2x-3}\)

b) Có \(\left|3x-2\right|+1=5\)

<=> \(\left|3x-2\right|=4\)

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

TH1: Thay x = 2 vào P, ta có:

P = \(\dfrac{-1}{2.2-3}=-1\)

TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:

P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)

c) Để P > 0

<=> \(\dfrac{-1}{2x-3}>0\)

<=> 2x - 3 <0

<=> x < \(\dfrac{3}{2}\) ( x khác 1)

d) P = \(\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)

<=> 2x - 3 = x² - 6

<=> x² - 2x - 3 = 0

<=> (x-3)(x+1) = 0

<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{39}{7}:\left\{x\cdot\dfrac{10}{13}+7.2\cdot\dfrac{257}{79}\right\}=\dfrac{15}{14}\)

\(\Leftrightarrow x\cdot\dfrac{10}{13}+\dfrac{9252}{395}=\dfrac{26}{5}\)

\(\Leftrightarrow x\simeq-23,69\)

b: TH1: x<1/2

Pt sẽ là 2-3x+1-2x=4

=>-5x+3=4

=>-5x=1

=>x=-1/5(nhận)

TH2: 1/2<=x<2/3

Pt sẽ là 2x-1+2-3x=4

=>1-x=4

=>x=-3(loại)

TH3: x>=2/3

Pt sẽ là 3x-2+2x-1=4

=>5x-3=4

=>5x=7

=>x=7/5(nhận)

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)

b) \(\left|2x+3\right|=x+2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)

Đính chính

Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)

\(\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{1}{3}\\2x-\dfrac{2}{3}=\dfrac{-1}{3}\end{matrix}\right.\left[{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

a)\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c)\(\left[{}\begin{matrix}x-2=2x-1\\-x+2=2x-1\end{matrix}\right.\left[{}\begin{matrix}x-2=2x-1\\-x+2=2x-1\end{matrix}\right.\left[{}\begin{matrix}x-2x=-1+2\\-x-2x=-1-2\end{matrix}\right.\left[{}\begin{matrix}-1x=1\\-3x=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=1:\left(-1\right)\\x=-3:\left(-3\right)\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

`#3107.101107`

`1/2x + 4/5 = 2x - 8/5`

`=> 1/2x - 2x = -4/5 - 8/5`

`=> -3/2x = -12/5`

`=> x = -12/5 \div (-3/2)`

`=> x = 8/5`

Vậy, `x = 8/5`

_____

`\sqrt{x} = 5`

`=> x = 5^2`

`=> x = 25`

Vậy, `x = 25`

___

`x^2 = 3`

`=> x^2 =  (+-\sqrt{3})^2`

`=> x = +- \sqrt{3}`

Vậy, `x \in {-\sqrt{3}; \sqrt{3}}.`

\(x=-17\)

P/s: Nhớ tick cho mình nha. Thanks bạn

`(2x+1)/3=(x-5)/2`

`=>2(2x+1)=3(x-5)`

`=>4x+2=3x-15`

`=>4x-3x=-15-2`

`=>x=-17`

Vậy `x=-17`

`3/(x+2) = 5/(2x+1)`

`=> 3(2x+1) = (x+2)5`

`=> 6x + 3 = 5x +10`

`=> 6x-5x=10-3`

`=>x=7`

a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)

\(\Rightarrow2x=\dfrac{4}{3}-\dfrac{-2}{3}\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)

\(\Rightarrow-5:x=\dfrac{5}{8}-\dfrac{-3}{8}\)

\(\Rightarrow-5:x=1\)

\(\Rightarrow x=-5\)

a) \(\dfrac{-2}{3}+2x=\dfrac{4}{3}\)

\(\Rightarrow2x=\dfrac{4}{3}-\dfrac{-2}{3}\)

\(\Rightarrow2x=\dfrac{6}{3}\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=\dfrac{2}{2}\)

\(\Rightarrow x=1\)

b) \(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)

\(\Rightarrow-5:x=\dfrac{-3}{8}-\dfrac{5}{8}\)

\(\Rightarrow-5:x=-\dfrac{8}{8}\)

\(\Rightarrow-5:x=-1\)

\(\Rightarrow x=-5:-1\)

\(\Rightarrow x=5\)

Mink sửa lại câu b nhé!

\(\dfrac{5}{8}-5:x=\dfrac{-3}{8}\)

\(\Rightarrow5:x=\dfrac{5}{8}-\dfrac{-3}{8}\)

\(\Rightarrow5:x=1\)

\(\Rightarrow x=5\)