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18 tháng 11 2018

Câu 1 ;

a) \(x^2-2x-15\)

\(x^2-5x+3x-15\)

\(x(x-5)+3(x-5)\)

\((x+3).(x-5)\)

b) \(xy+\frac{1}{3}y-\frac{1}{4}x-\frac{1}{12}\)

\((x+\frac{1}{3})y-\frac{1}{4}(x+\frac{1}{3})\)

\((x-\frac{1}{4}).(x+\frac{1}{3})\)

Câu 2 : 

\(A=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+1994\)

=> \(A=x^3+1+x-x^3+1+1994\)

=> \(A=1+x+1+1994\)

=> \(A=x+1996=-1995+1996=1\)

17 tháng 10 2019

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

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

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

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

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

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

Giờ chỉ cần thế x, y vô nữa là xong nhé.

17 tháng 10 2019

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

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

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

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

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

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

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

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

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

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

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

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

Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :

\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)

16 tháng 12 2017

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

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

Tại   x = \(\frac{1}{2}\)thì:

             A = \(\frac{2.\frac{1}{2}-4}{\frac{1}{2}+2}\)\(=\)\(\frac{-3}{\frac{5}{2}}\)\(=\)\(\frac{-6}{5}\)

15 tháng 12 2021

\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)

8 tháng 1 2022

mk mới lớp 5 nên ko bt

6 tháng 7 2017

a. \(=4x^2-4xy+y^2+4x^2-4xy+y^2=8x^2+2y^2\)

\(=8.\left(\frac{1}{21}\right)^2+4.\left(-0.3\right)^2=\frac{4169}{11025}\)

b, \(=\left(\frac{1}{7}xy+7yz+\frac{1}{7}xy-7yz\right)\left(\frac{1}{7}xy+7yz-\frac{1}{7}xy+7yz\right)\)

\(=\frac{2}{7}xy.14yz=4xy^2z=4.2.\left(0,25\right)^2.\left(-4\right)=-2\)

7 tháng 12 2018

các bạn giải nhanh cho mình nhé vì mình đang cần gấp

7 tháng 12 2018

Mình nghĩ bạn viết hơi sai đề bài.

\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)

Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)

\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)

Khi đó:

 \(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)

\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)