a: \(4x^2\left(3x^{n+1}-2x^n\right)\)

\(=12x^{n+3}-8x^{n+2}\)

b: \(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)

\(=2x^{2n}\)

c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)=x^{6n}-y^{6n}\)

d: \(=4^n\cdot4-3\cdot4^n=4^n\)

a: \(=12x^{n+2}+4x^2-8x^{n+2}\)

\(=4x^{n+2}+4x^2\)

b: \(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)

\(=2x^{2n}\)

c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)

\(=x^{6n}-y^{6n}\)

d: \(=4^n\cdot4-3\cdot4^n=4^n\)

Bài 1:

Ta có:

\(a+b+c=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc\right)=0\\ \Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow a^3+b^3+c^3=3abc\left(dpcm\right)\)

hơi nhiều nhỉ

Sao bạn đăng nhiều thế !

hoa mắt thì làm sao giải cho bạn được

Bài 1:

(2x -1) (3y + 2) = 12b

\(x=\frac{12b+3y+2}{2\left(3y+2\right)}\)

\(y=\frac{2\left(6b-2x+1\right)}{3\left(2x-1\right)}\)

(4x + 1) (2y-3) = -81

\(x=-\frac{y+39}{2\left(2y-3\right)}\)

\(y=\frac{3\left(2x-13\right)}{4x+1}\)

Mk lm câu b bài 2 há!

b, ( 8x - 3 )( 3x + 2 ) - ( 4x + 7 )( x + 4 ) = ( 2x +1 )( 5x - 1) =- 33

Pt <=> 3x ( 8x - 3 ) + 2( 8x- 33) - ( x ( 4x + 7) ) + ( 2x + 1) - 5x ( 2x + 1) + 33 = 0

<=> 24x² - 9x + 16x - 6 - ( 4x² + 7x + 16x + 28) + 2x + 1 - 10x² - 5x + 33 = 0

<=> 24x² - 9x + 16x - 6 - 4x² - 7x - 16x - 28 + 2x + 1 - 10x² - 19x = 0 <=> x ( 10x - 19) = 0 

=> \(\orbr{\begin{cases}x=0\\10x-19=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{19}{10}\end{cases}}\) 

^^ Ok con tê tê!

a) =2x^3-10x^2-2x+3x^2-x

=2x^3-7x^2-3x

b) -10x^4y^2z^2+35x^3y^2z^2+4x^4y^2z^2+4x^3y^2z^2

=-6x^4y^2z^2+39x^3y^2z^2