2x^2 + 3x - ( x^2 + 3x + 2 ) = 6 

2x^2 + 3x - x^2 - 3x - 2 = 6 

x^2 - 2 = 6 

x^2 = 8 

x = +- căn 8 

Ta có : x(2x + 3) - (x + 2)(x + 1) = 6

=> 2x² + 3x - (x² + 3x + 2) = 6

=> 2x² + 3x - x² - 3x - 2 = 6

=> x² = 8

=> \(x=\pm\sqrt{8}\)

là tự đi mà tìm hiểu nhé

Bài 1.

a) ( 3x + 4y )² = ( 3x )² + 2.3x.4y + ( 4y )² = 9x² + 24xy + 16y²

b) ( x² + 1 )² = ( x² )² + 2.x².1 + 1² = x⁴ + 2x² + 1

c) ( 3 - 2y )² = 3² - 2.3.2y + ( 2y )² = 9 - 12y + 4y²

d) ( xy² - 2 )² = ( xy² )² - 2.xy².2 + 2² = x²y⁴ - 4xy² + 4

Bài 2.

a) x² - 9 = x² - 3² = ( x - 3 )( x + 3 )

b) 25 - 4y² = 5² - ( 2y )² = ( 5 - 2y )( 5 + 2y )

c) 9x⁴ - 4y² = ( 3x² )² - ( 2y )² = ( 3x² - 2y )( 3x² + 2y )

d) ( x + 1 )² - y² = ( x - y + 1 )( x + y + 1 )

B1:

a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)

b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)

c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)

d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)

B2:

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

b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)

c) \(9x^4-4y^2=\left(3x^2\right)^2-\left(2y\right)^2=\left(3x^2-2y\right)\left(3x^2+2y\right)\)

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

ko bt tự làm đê!!!!
 

27(1-x)(x²+x+1)+81x(x-1)

=(27-27x)(x²+x+1)+81x²-81x

=27x²-27x³+27x-27x²+27-27x+81x²-81x

=-27x³+81x²-81x+27

\(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)

\(=-27\left(x-1\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)

\(=-27\left(x-1\right)\left(x^2+x+1-3x\right)\)

\(=-27\left(x-1\right)\left(x^2-2x+1\right)\)

\(=-27\left(x-1\right).\left(x-1\right)^2\)

\(=-27.\left(x-1\right)^3\)

Có nguyên tố khối à ? Hay bạn viết ngược vậy ?

\(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{9}{20}\)

Ta có : \(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{9}{20}\)(Vì a + b = 9 ; ab = 20)

Vậy \(\frac{1}{a}+\frac{1}{b}=\frac{9}{20}\)