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 2019

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

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

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

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

29 tháng 11 2019

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

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

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

14 tháng 2 2020

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

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

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)

8 tháng 7 2016

\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\)

\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right):\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)

\(P=\left(\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\right):\frac{2x^2+y+2}{x+1}\)

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

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

 

8 tháng 7 2016

thanks hihi

8 tháng 7 2016

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

      \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

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

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

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

       \(=\left[\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}:\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)

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

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

Bài 1:

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

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

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

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

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

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

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

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

\(=\frac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

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

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

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

c) Ta có: \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}\)

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

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

d) Ta có: \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)

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

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

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

Bài 2:

a) Ta có: \(\frac{4x+1}{2}-\frac{3x+2}{3}\)

\(=\frac{3\left(4x+1\right)}{6}-\frac{2\left(3x+2\right)}{6}\)

\(=\frac{12x+3-6x-4}{6}\)

\(=\frac{6x-1}{6}\)

b) Ta có: \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)

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

\(=\frac{x^2-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)

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

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

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

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

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

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

\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x+1\right)\left(x-1\right)}-\frac{2\cdot2\cdot\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)

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

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

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

d) Ta có: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}\)

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

f) Ta có: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)

22 tháng 3 2020

.các bác giúp em với ạ,em cảm ơn trc ạ

22 tháng 3 2020

bài 1 :

1, 

a, x^2 - xy = x(x - y)

b, x^2 + 2xy + y^2 - 4

= (x + y)^2 - 2^2

= (x + y + 2)(x + y - 2)

2,

(2x-1)(2x+1)+4x(1-x)

= 4x^2 - 1 + 4x - 4x^2

= 4x - 1

3,  x^2 - 6x + 5 = 0

<=> x^2 - x - 5x + 5 = 0

<=> x(x - 1) - 5(x - 1) = 0

<=> (x - 5)(x - 1) = 0

<=> x = 5 hoặc x = 1