bài 1.Rút gọn :
A=\(\frac{2a^3b^5}{3a^3b^2}\)
B=\(\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
K
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DT
1
9 tháng 12 2018
\(\frac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\frac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\frac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\frac{x-y+z}{x-y-z}\)
LN
1
TN
27 tháng 5 2016
A=\(\frac{2xy-x^2+z^2-y^2}{x^2+z^2-y^2+2xz}\)=\(\frac{z^2-\left(x^2-2xy+y^2\right)}{\left(x^2+2xz+z^2\right)-y^2}\)=\(\frac{z^2-\left(x-y\right)^2}{\left(x+z\right)^2-y^2}\)=\(\frac{\left(z+x-y\right)\left(z-x+y\right)}{\left(x+z-y\right)\left(x+z+y\right)}\)=\(\frac{\left(z-x+y\right)}{\left(x+z+y\right)}\)
G
11 tháng 7 2019
Ta có: \(\hept{\begin{cases}xy+x+y=1\\yz+y+z=3\\xz+x+z=7\end{cases}}\Rightarrow\hept{\begin{cases}xy+x+y+1=2\\yz+y+z+1=4\\xz+x+z+1=8\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=2\\\left(y+1\right)\left(z+1\right)=4\\\left(x+z\right)\left(z+1\right)=8\end{cases}}\)
Nhân theo vế:
\(\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=64\Rightarrow\orbr{\begin{cases}\left(x+1\right)\left(y+1\right)\left(z+1\right)=8\\\left(x+1\right)\left(y+1\right)\left(z+1\right)=-8\end{cases}}\)
Thay vào từng trường hợp tìm x;y;z
TV
0
LQ
0
3 tháng 8 2019
\(\frac{x}{5}=\frac{y}{6}\Rightarrow\frac{x}{20}=\frac{y}{24};\frac{y}{8}=\frac{z}{7}\Rightarrow\frac{y}{24}=\frac{z}{21}\Rightarrow\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\Rightarrow x=60;y=72;z=63\)
PT
1
TH
29 tháng 6 2017
\(\frac{2xy-x^2+z^2-y^2}{-x^2+y-z^2+2xz}\)
\(=\frac{-\left[\left(x^2-2xy+y^2\right)-z^2\right]}{-\left[\left(x^2-2xz+z^2\right)-y\right]}\)
\(=\frac{-\left[\left(x-y\right)^2-z^2\right]}{-\left[\left(x-z\right)^2-y\right]}\)
\(=\frac{-\left(x-y-z\right)\left(x-y+z\right)}{-\left(x-z\right)^2+y}\)
Ta có: \(A=\frac{2a^3b^5}{3a^3b^2}=\frac{2b^3}{3}\)
Ta có:
\(B=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y+z\right)\left(x+y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
A= \(\frac{2b^3}{3}\)
B= \(\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+z+y\right)\left(x+z-y\right)}=\frac{x+y-z}{x+z-y}\)