x² - 3x - 2x² + 2x = -x² + 5x

x² - 3x - 2x² + 2x + x² - 5x = 0

(x² - 2x² + x²) + (-3x + 2x - 5x) = 0

-6x = 0

x = 0

giải dùm bn thoi chứ h mk k cần vì mk có kinh nghiem ke hứa nhieu

k bao gio giu loi hua

2x -3x +5x  = 3/3 - 4/4 + 12/5

4x = 1-1 + 12/5 

x = 3/5

b) \(\Leftrightarrow5x^2-6x+1=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow x=\frac{1}{5}\) hoặc x = 1

c) \(\Leftrightarrow x^2+4x-21-x^2-4x+5=0\Leftrightarrow-16=0\) (vô lí) => PT vô nghiệm

d) \(\Leftrightarrow x^2+3x-10=0\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\)x = 2 hoặc x = -5

e) \(\Leftrightarrow x\left(x-2\right)=0\)<=> x = 0 hoặc x = 2

a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)

\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)

b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)

\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)

c) \(14-\left|\frac{3x}{2}-1\right|=9\)

\(\Rightarrow\left|\frac{3x}{2}-1\right|=14-9=5\)

\(\Rightarrow\orbr{\begin{cases}\frac{3x}{2}-1=5\\\frac{3x}{2}-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{3x}{2}=6\\\frac{3x}{2}=-4\end{cases}}\Rightarrow\orbr{\begin{cases}3x=12\\3x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{-8}{3}\end{cases}}\)

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(8-5x)=0`

`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\) 

Vậy `S={-3,8/5}`

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(-5x+8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)

Vậy `S={-3;8/5}`.

\(\left(x+3\right)\left(x^2-5x+8\right)=\left(x+3\right).x^2\\ \Leftrightarrow\left(x+3\right)\left(x^2-5x+8\right)-x^2\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-5x+8-x^2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(8-5x\right)=0\\\Leftrightarrow x+3=0 \)

hoặc \(8-5x=0\)

\(x+3=0\\ \Leftrightarrow x=-3\)

\(8-5x=0\\ \Leftrightarrow-5x=-8\\ \Leftrightarrow5x=8\\ \Leftrightarrow x=\dfrac{8}{5}\)

Vậy pt có tn là \(S=\left\{-3;\dfrac{8}{5}\right\}\)