Sao ảnh đại diện của bạn giống mình thế?

a/ 

\(\frac{a+b+c}{\left(a+b\right)^2-c\left(a+b\right)}.\frac{2a+2b}{a^2+2ab-c^2+b^2}\)

\(=\frac{a+b+c}{\left(a+b\right)\left(a+b-c\right)}.\frac{2\left(a+b\right)}{\left(a+b+c\right)\left(a+b-c\right)}\)

\(=\frac{2}{\left(a+b-c\right)^2}\)

b/ \(\frac{3x+3y}{x^2+y^2-2xy}:\frac{6x+6y}{ax-by+bx-ay}\)

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

\(=\frac{a-b}{2\left(x-y\right)}\)

a. 2x(x + y) - y(y + 2x) = 2x² + 2xy - y² - 2xy = 2x² - y²

b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

Phần c nản quá.

a) 2x(x + y) - y(y + 2x) 

= 2x² + 2xy - y² - 2xy

= 2x² - y²

b) \(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

c) \(\frac{x^3-4x^2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{x-1}\)

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

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

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

Bài 1:

a)\(-x^2\left(3x^3-2x+\frac{1}{2}\right)=-3x^5+2x^3-\frac{x^2}{2}\)

b) Sửa đề: \(\left(2xy-y^2+x\right)\frac{2}{3}x^3y=\frac{4}{3}x^4y^2-\frac{2}{3}x^3y^3+\frac{2}{3}x^4y\)

Không sửa đề: \(\left(2xy-y2+x\right)\frac{2}{3}x^3y=\frac{4}{3}x^4y^2-\frac{4}{3}x^3y^2+\frac{2}{3}x^4y\)

Bài 2:

a)\(3x\left(12x-5\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-15x-36x^2+27x=30\)

\(\Leftrightarrow12x=30\Leftrightarrow x=\frac{5}{2}\)

b) \(x\left(7-2x\right)+2x\left(x-4\right)=15\)

\(\Leftrightarrow7x-2x^2+2x^2-8x=15\)

\(\Leftrightarrow-x=15\Leftrightarrow x=-15\)

Hok tốt

Bài 1:

a) \(x.\left(x^2-2xy+1\right)=x^3-2x^2y+x\)

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

Bài 2:

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

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

c) Đề sai.