Chia ra 2 trường hợp nha bạn 

a/ TH1 : 2x + 1 = 2x + 1

=> 2x - 2x = 1 - 1 

=> 0x = 0

vậy có vô số x thỏa mãn biểu thức trên

TH2 : 2x + 1 = -2x - 1

=> 2x + 2x = -1 - 1

=> 4x = -2 => x = -1/2

b/ => I 2-x I = 6

TH1 : 2 - x = 6 =>x = -4

TH2 : 2 - x = -6 => x = -8

c/ TH1: 3x - 1 = 1 - 3x 

=> 3x + 3x = 1 + 1

=> 6x = 2

=> x = 1/3

TH2 : 3x - 1 = 3x - 1 

=> 3x - 3x = 1 - 1

=> 0x = 0

Vậy có vô số x thỏa mãn biểu thức trên

a) x⁴ + 3x² + 4

= [ ( x² )² + 4x² + 2² ] - x²

= ( x² + 2 )² - x²

= ( x² + 2 - x² ).( x² + 2 + x² )

= 2( 2x² + 2 )

b) x⁴ + 5x² + 9

= [ ( x² )² + 6x² + 3² ] - x²

= ( x² + 3 )² - x²

= ( x² + 3 - x² )( x² + 3 + x² )

= 3( 2x² + 3 )

=8x^2+12x=4x(2x+3)

\(x^2+7x^2+12x=8x^2+12x=4x\left(2x+3\right)\)

mk viết đáp án, ko biết biến đổi ib mk

a)  \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)

b)    \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)

c)   \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)

d)   \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)

\(\text{a) }x^3y^3+x^2y^2+4\)

\(=x^3y^3+2x^2y^2-x^2y^2+4\)

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

\(=x^2y^2\left(xy+2\right)-\left(xy+2\right)\left(xy-2\right)\)

\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)

\( {c)}\)\(x^4+x^3+6x^2+5x+5\)

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

\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2+5\right)\)

\({d)}\)\(x^4-2x^3-12x^2+12x+36\)

\(=\left(x^4-2x^3-6x^2\right)-\left(6x^2-12x-36\right)\)

\(=x^2\left(x^2-2x-6\right)-6\left(x^2-2x-6\right)\)

\(=\left(x^2-2x-6\right)\left(x^2-6\right)\)

Câu b sai đề thì phải ah

a) x² - 4x + 2 = (x² - 4x + 4) - 2 = (x - 2)² - 2 = \(\left(x-2+\sqrt{2}\right)\left(x-2-\sqrt{2}\right)\)

b)  x² - 12x + 11 = x² - x - 11x + 11 = x(x - 1) - 11(x - 1) = (x - 1)(x - 11)

c) 3x² + 6x - 9 = 3x² - 3x + 9x - 9 = 3x(x - 1) + 9(x - 1) = (3x + 9)(x - 1) = 3(x + 3)(x - 1)

d) 2x² - 6x + 2 = 2(x² - 3x + 1) = 2(x² - 3x + 9/4 - 5/4) = 2[(x - 3/2)² - 5/4] = \(2\left(x-\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\left(x-\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\) 

1. 

a) \(x^2-4x+2=\left(x^2-4x+4\right)-2=\left(x-2\right)^2-2=\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)\)

b) \(x^2-12x+11=\left(x^2-12x+36\right)-25=\left(x-6\right)^2-5^2=\left(x-6-5\right)\left(x-6+5\right)=\left(x-11\right)\left(x-1\right)\)

c) \(3x^2+6x-9=3\left(x^2+2x-3\right)=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)

d) \(2x^2-6x+2=2\left(x^2-3x+1\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]\)

\(=2\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)

=5400

\(\text{6.75.1.25-0.675.120-6.75.13}\)

\(=5400\)

Bạn nhé !!! k với nha !!!

=5400

6.75*3=20.25

(7.14*3)*3=64.26

(9.26*2)*9=166.68