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RB
1
11 tháng 5 2022
\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(b,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
HB
1
17 tháng 2 2023
\(A=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-b\right)}=\dfrac{2a\left(x-1\right)^2}{5b\left(1-b\right)}\)
\(B=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
6 tháng 12 2016
B=\(\frac{5\left(x-y\right)-3\left(x-y\right)}{10\left(x-y\right)}\)
B=\(\frac{\left(x-y\right)\left(5-3\right)}{10\left(x-y\right)}\)
B= \(\frac{\left(x-y\right)2}{10\left(x-y\right)}\)
B= 5
vậy B=5
TN
13 tháng 8 2017
1) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{-5b\left(x^2-1\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
2) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+x+3x+3}{2\left(x+3\right)}=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{2\left(x+3\right)}\)\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3)\(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)Học tốt nha you<3
p/s: tớ ko bk đã rút gọn hết chưa:(
13 tháng 8 2017
1, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
2, \(\dfrac{x^2+4x+3}{2x+6}\)
\(=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
5, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
22 tháng 9 2019
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
Ta thay : \(x=0;y=2009;z=2010\) ta được :
\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)
Chúc bạn học tốt !!!
T
22 tháng 9 2019
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)
Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :
\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)
\(\frac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=-\frac{2a\left(x-1\right)^2}{5b\left(x+1\right)\left(x-1\right)}\)
\(=-\frac{2a\left(x-1\right)}{5b\left(x+1\right)}\)