K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 11 2022

\(=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

29 tháng 11 2022

Sửa đề: \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\)

\(=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

14 tháng 12 2021

\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)

11 tháng 12 2017

\(a.\)

\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)

\(=\left(x-5\right)\left(x+5\right).\dfrac{3x-7}{2\left(x+5\right)}\)

\(=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}\)

\(=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)

\(b.\)

\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)

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

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

\(=\dfrac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}\)

\(=\dfrac{x}{3\left(x-1\right)}\)

11 tháng 12 2017

\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}=\dfrac{5x\left(x+1\right)\left(x-1\right)}{15\left(x-1\right)^2\left(x+1\right)}=\dfrac{x}{3\left(x-1\right)}\)\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)

1 tháng 4 2018

Hỏi đáp ToánChúc bạn học tốt!

8 tháng 12 2017

Phép cộng các phân thức đại số

1 tháng 4 2018

2xyz+4xyz-\(\frac{1}{2}\) xyz

=(2+4-\(\frac{1}{2}\) )(xxxyyyzzz)

=3.5xy3z3

9 tháng 12 2018

\(\dfrac{18x}{x^3-9x}-\dfrac{2-x}{x+3}+\dfrac{3}{3-x}\)

=\(\dfrac{18x}{x\left(x^2-9\right)}-\dfrac{\left(2-x\right)\left(x-3\right)x}{\left(x+3\right)\left(x-3\right)x}-\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)x}\)

=\(\dfrac{18x-\left(2-x\right)\left(x-3\right)x-3x\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)

=\(\dfrac{18x-\left(2x-6-x^2+3x\right)x-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)

=\(\dfrac{18x-2x^2+6x+x^3-3x^2-3x^2-9x}{x\left(x+3\right)\left(x-3\right)}\)

=\(\dfrac{x^3-8x^2+15x}{x\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{x\left(x^2-8x+15\right)}{x\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{x^2-3x-5x+15}{x\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{x\left(x-3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{\left(x-5\right)}{x+3}\)

8 tháng 12 2018

BẠN CŨNG CHƠI MINECRAFT AK

14 tháng 12 2021

undefined

4 tháng 12 2017

\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)

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

\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)

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

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

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

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

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

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

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

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

4 tháng 12 2017

\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)

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

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

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

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

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