\(\left(x-3\right)^2-5\left(x-2\right)+5=0\\ \Leftrightarrow x^2-6x+9-5x+10+5=0\\ \Leftrightarrow x^2-11x+24=0\\ \Leftrightarrow\left(x-8\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)

\(\left(2x-1\right)^2-3\left(x-2\right)\left(x+2\right)-25=0\\ \Leftrightarrow4x^2-4x+1-3\left(x^2-4\right)-25=0\\ \Leftrightarrow4x^2-4x-24-3x^2+12=0\\ \Leftrightarrow x^2-4x-12=0\\ \Leftrightarrow\left(x-6\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

a: Ta có: \(\left(x-3\right)^2-5\left(x-2\right)+5=0\)

\(\Leftrightarrow x^2-6x+9-5x+10+5=0\)

\(\Leftrightarrow x^2-11x+24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)

b: Ta có: \(\left(2x-1\right)^2-3\left(x-2\right)\left(x+2\right)-25=0\)

\(\Leftrightarrow4x^2-4x+1-3x^2+12-25=0\)

\(\Leftrightarrow x^2-4x-12=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)

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

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

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

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

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

=>x+2=0

=>x=-2

\(\Leftrightarrow x^3-\left(m-1\right)x^2-\left(m-1\right)x-2x^2+2\left(m-1\right)x+2m-2=0\)

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

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

Mình làm ý b,c thôi a tương tự b 

b) 5x^2 - 13x = 0 

=> x(5x - 1 3) = 0 

=> x = 0 hoặc 5x - 13 = 0 

=> x = 0 hoặc x = 13/5 

b) x + 1 = ( x+1  )^2 

=> (x + 1 )^2 - (x+ 1) =  0 

=> (x +1  )( x + 1  - 1 ) = 0 

=> x(x + 1 ) = 0 

=> x=  0 hoặc x + 1 = 0 

=> x = 0 hoặc x = -1 

a, x+5x²=0

<=>x(1+5x)=0

<=>x=0 hoặc 1+5x=0

<=>x=0 hoặc x=-1/5

b, 5x²-13x=0

<=>x(5x-13)=0

<=>x=0 hoặc 5x-13=0

<=>x=0 hoặc x=13/5

c, x+1=(x+1)²

<=>(x+1)²-(x+1)=0

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

<=>x(x+1)=0

<=>x=0 hoặc x+1=0

<=>x=0 hoặc x=-1

\(\left(3x^2-1\right)-\left(x+3\right)^2\)

\(=\left(3x^2-1-x-3\right)\left(3x^2-1+x+3\right)\)

\(=\left(3x^2-x-4\right)\left(3x^2+x+2\right)\)

\(=\left(3x^2+3x-4x-4\right)\left(3x^2+x+2\right)\)

\(=\left(x+1\right)\left(3x-4\right)\left(3x^2+x+2\right)\)

ai nhanh và đúng mik tích cho

Cần lao vi tiên thủ vô vi thực DB thực CUC

Tự làm đi bạn êi :D

a)a) x^2-16
<=> x^2-4^2
<=> (x-4)(x+4)

câu 1 (x-4)*(x+4)

câu 2 x=-2/3, x=5

câu 1: 
x²-16=x²-4²=(x+4)(x-4)
câu 2:
3x(x-5)+2(x-5)=0
(3x+2)(x-5)=0
TH1: 3x+2=0       TH2: x-5=0
3x=-2                               x=5
x=-2/3
vậy x\(\in\)\(\left\{\dfrac{-2}{3};5\right\}\)

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)