a) Ta có: \(\left|2x-1\right|=\left|2x+3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x+3\left(loại\right)\\2x-1=-2x-3\end{matrix}\right.\Leftrightarrow2x+2x=-3+1\)

\(\Leftrightarrow4x=-2\)

hay \(x=-\dfrac{1}{2}\)

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

\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)

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

\(=\left(x+1\right)\cdot\left(-1\right)\)

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

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

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)

\(=x^3-1-x^3-8+12x-12\)

\(=-21+12x\)

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

\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)

\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)

\(=0\)

Chỗ dấu bằng thứ hai sai nên bạn làm cũng chưa đúng

x^6 -y^6 = (x^2-y^2)(x^4 +x^2 .y^2 + y^4)

Bạn hiểu ra chỗ sai của mình chưa.Chúc bạn học tốt.

cho minh xin de

Phân đa thức thành nhân tử

a) \(\left(x+2\right)\left(x^2-4x+4\right)-\left(x^3+2x^2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(4-4x\right)=5\)

\(\Leftrightarrow4x-4x^2+8-8x=5\)

\(\Leftrightarrow-4x^2-4x+3=0\)

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

\(\Leftrightarrow4x^2-2x+6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

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

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

Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)

b) \(6x^2-6x\left(-2+x\right)=36\)

\(\Leftrightarrow6x^2+12x-6x^2=36\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(x+2\right)^2+\left(x-3\right)^2-2\left(x-1\right)\left(x+1\right)=9\)

\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)

\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)

\(\Leftrightarrow15-2x=9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy x = 3

d) \(\left(x+5\right)^2-9=0\)

\(\Leftrightarrow\left(x+5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)

Vậy x ={-2; -8}

e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)

\(\left(x-2\right)^3-x^3+6x^2=7\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)

\(\Leftrightarrow12x-8=7\)

\(\Leftrightarrow12x=15\)

\(\Leftrightarrow x=\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

#Chúc bạn học tốt!

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

a) \(\left(3x-1\right)^2-3x\left(x-5\right)=21\)

\(\Leftrightarrow9x^2-6x+1-3x^2+15x=21\)

\(\Leftrightarrow6x^2+9x-20=0\)

\(\Leftrightarrow x\in\left\{-\sqrt{\frac{\sqrt{561}+9}{12}};\sqrt{\frac{\sqrt{561}-9}{12}}\right\}\)

b) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-2\right)=36\)

\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)

\(\Leftrightarrow8x+76=36\)

\(\Leftrightarrow8x=-40\)

\(\Leftrightarrow x=-5\)

Đặng Phương Thảo Bạn đến cùng thời gian với vài nick ra đi,haizz.

uk Minh Triều cãi nhau vs tụi mk ra đi rồi 

\(=3x^2\left(x^2-1\right)+\left(x^8-3x^4+3x^2-1\right)-\left(x^8-1\right)\)

\(=3x^4-3x^2+x^8-3x^4+3x^2+1-x^8+1\)

\(=2\)

=2 nha ban

(con cach lam ban nhan dang thuc len rui rut gon lai)

Cm: Ta có: 

a) A = x² - 8x + 20 = (x² - 8x + 16) + 4 = (x - 4)² +  4 > 0 \(\forall\) x(vì (x - 4)² \(\ge\)0 \(\forall\)x ; 4 > 0)

=> A luôn dương với mọi x

b) B = 4x² - 12x + 11 = [(2x)² - 12x + 9] + 2 = (2x - 3)² + 2 > 0 \(\forall\)x (vì (2x - 3)² \(\ge\)0 \(\forall\)x; 2 > 0)

=> B luôn dương với mọi x

c) C = x² - x + 1 = (x² - x + 1/4) + 3/4 = (x -  1/2)² + 3/4 > 0 \(\forall\)x (vì (x - 1/2)² \(\ge\)0 \(\forall\)x; 3/4 > 0)

=> C luôn dương với mọi x

* Tìm x

3(x + 2)² + (2x - 1)² - 7(x + 3)(x - 3) = 36

=> 3(x² + 4x + 4) + 4x² - 4x + 1 - 7(x² - 9) = 36

=> 3x² + 12x + 12 + 4x² - 4x + 1 - 7x² + 63 = 36

=> 8x + 76 = 36

=> 8x = 36 - 76

=> 8x = -40

=> x = -40 : 8 = -5