b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5

a) <=> \(2x^2-8x+3x-12+x^2-7x+10=3x^2-5x-12x+20\)

<=> \(2x^2-8x+3x-12+x^2-7x+10-3x^2+5x+12x-20=0\)

<=> \(5x-22=0\)

<=> \(5x=22\)

<=> \(x=\frac{22}{5}\)

b) <=> \(24x^2-9x+16x-6-4x^2-7x-16x-28=10x^2+5x-2x-1\)

<=> \(24x^2-9x+16x-6-4x^2-7x-16x-28-10x^2-5x+2x+1=0\)

<=> \(10x^2-19x-33=0\)

<=> \(10x^2-30x+11x-33=0\)

<=> \(10x\left(x-3\right)+11\left(x-3\right)=0\)

<=> \(\left(x-3\right)\left(10x+11\right)=0\)

<=> \(x=3;x=-\frac{11}{10}\)

d) \(x\left(x+1\right)-x-1=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

<=> 2x(x + 2) = 1

Vì 1 chỉ có tích là 1 . 1 nên:

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

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

a,\(\Leftrightarrow\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)-17=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x-17=0\)

\(\Leftrightarrow9x-10=0\)

\(\Leftrightarrow x=\frac{10}{9}\)

b,\(\Leftrightarrow x^3+8-x^3+2x-15=0\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{7}{2}\)

c,\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+1-15=0\)

\(\Leftrightarrow45x=14\)

\(\Leftrightarrow x=\frac{14}{45}\)

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

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

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

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

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

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)