a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

Tìm x:

a) \(x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow x^2-x-30=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=6\\ x=-5 \end{array} \right.\)

b) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow2-4x=0\)

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

c) \(x^2\left(x^2+4\right)-x^2-4=0\)

\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)

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

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

a) \(5\left(x+7\right)-12x=15\)

\(5x+35-12x=15\)

\(-7x=15-35\)

\(-7x=-20\)

\(x=\frac{20}{7}\)

vay \(x=\frac{20}{7}\)

b) \(x^2-25-\left(x+5\right)=0\)

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

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

\(\left(x+5\right)\left(x-5-1\right)=0\)

\(\left(x+5\right)\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)

vay \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)

c) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\left(2x-1\right)\left(2x-1\right)-\left(\left(2x\right)^2-1^2\right)=0\)

\(\left(2x-1\right)\left(2x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)

\(\left(2x-1\right)\left(2x-1-2x-1\right)=0\)

\(-2.\left(2x-1\right)=0\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow x=\frac{1}{2}\)

vay \(x=\frac{1}{2}\)

d) \(x^2.\left(x^2+4\right)-x^2-4=0\)

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

\(\left(x^2-1\right)\left(x^2+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=1\\x^2=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1hoacx=-1\\kotontai\end{cases}}\)

vay \(x=1\)hoac \(x=-1\)

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

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

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

vậy ...

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

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

a: Ta có: \(x^4-2x^3+5x^2-10x=0\)

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

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

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

b:Ta có: \(\left(3x+5\right)^2=\left(2x-2\right)^2\)

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

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

\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

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