\(\frac{1}{2}X+\frac{1}{3}Y\))\(^2\)-  ( 2X+...">
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2 tháng 3 2020
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2 tháng 3 2020

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

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

c. \(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}=\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)

d. \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}=\frac{xy}{2x-y}-\frac{1-x^2}{2x-y}=\frac{xy-1+x^2}{2x-y}\)

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

Bài 1:

a) Ta có: \(\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right):\frac{16x}{4x^2+4xy+y^2}\)

\(=\left(\frac{\left(2x+y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}+\frac{2\cdot\left(2x+y\right)\left(2x-y\right)}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}+\frac{\left(2x-y\right)^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\right)\cdot\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{\left(4x\right)^2}{\left(2x-y\right)^2}\cdot\frac{1}{16x}\)

\(=\frac{16x^2}{16x\cdot\left(2x-y\right)^2}\)

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

b) Ta có: \(\frac{3}{3x+3}+\frac{10}{5-5x}+\frac{5x-1}{x^2-1}\)

\(=\frac{1}{x+1}-\frac{2}{x-1}+\frac{5x-1}{x^2-1}\)

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

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

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

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

\(=\frac{4}{x+1}\)

c) Ta có: \(A=\left(x^4-x^2+2x-1\right):\left(x^2+x-1\right)-\left(x^2-x\right)\)

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

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

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

\(=x^2-x+1-x^2+x\)

=1

1 tháng 12 2019

1. Ta có:

\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)

\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)

\(=\frac{2}{x}-\frac{1}{x+2014}\)

\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)

\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)

1 tháng 12 2019

2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1

b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)

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

A = \(x-1+x+1-3\)

A = \(2x-3\)

c) Với x = 3 => A = 2.3 - 3 = 3

c) Ta có: A = -2

=> 2x - 3 = -2

=> 2x = -2 + 3 = 1

=> x= 1/2

29 tháng 11 2019

giúp mk với

29 tháng 11 2019

a) \(\frac{x-1}{2x}+\frac{2x+1}{2x}+\frac{1-5x}{6x}\)

\(=\frac{3x-3}{6x}+\frac{6x+3}{6x}+\frac{1-5x}{6x}\)

\(=\frac{3x-3+6x+3+1-5x}{6x}\)

\(=\frac{4x+1}{6x}\)

24 tháng 7 2017

a) \(\frac{x+1}{2x+6}\)+\(\frac{2x+3}{x\left(x+3\right)}\)

\(\frac{x+1}{2\left(x+3\right)}\)\(\frac{2x+3}{x\left(x+3\right)}\)

\(\frac{x\left(x+1\right)}{2x\left(x+3\right)}\)\(\frac{2\left(2x+3\right)}{2x\left(x+3\right)}\)

\(\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)

\(\frac{x^2+5x+6}{2x\left(x+3\right)}\)

\(\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)

\(\frac{x+2}{2x}\)

b) \(\frac{x-1}{x}\)\(\frac{x+2}{2}\)

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

\(\frac{2x-2+x^2+2x}{2x}\)

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

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

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

\(\frac{x^3+xy^2-x^2y-y^3-x^3-xy^2-xy^2-y^3+2x^3+2x^2y-2x^2y+2xy^2}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

\(\frac{2x^3+xy^2-x^2y-2y^3}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

\(\frac{\left(2x^3-2y^3\right)-\left(x^2y-xy^2\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

\(\frac{2\left(x-y\right)\left(x^2+xy+y^2\right)-xy\left(x-y\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

\(\frac{\left(x-y\right)\left(2x^2+2xy+2y^2-xy\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

\(\frac{2x^2+xy+2y^2}{\left(x+y\right)\left(x^2+y^2\right)}\)

e) = \(\frac{3x^2-6xy+3y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(\frac{3\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

=\(\frac{3x-3y}{x^2+xy+y^2}\)

( Mình bận rồi, lát làm câu d nhé)