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

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

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

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

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

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 !!!

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

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

Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :

\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)

Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

Bài 1 : 

\(N=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)

hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)

mà \(xyz=2\Rightarrow-xyz=-2\)

hay N nhận giá trị -2 

Bài 2 : 

\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)

hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)

hay biểu thức trên nhận giá trị là 24 

c, Ta có : \(a-b=3\Rightarrow a=3+b\)

hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)

\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi 

1.Ta có:\(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)

2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)

Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)

Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)

Vậy....

3r3reR

\(\orbr{\begin{cases}y=\frac{3}{x}\\z=\frac{4}{x}\end{cases}\Rightarrow\frac{12}{x^2}=6\Rightarrow x^2=2}\)

\(\orbr{\begin{cases}x=\frac{3}{y}\\z=\frac{6}{y}\end{cases}\Rightarrow\frac{18}{y^2}=4\Rightarrow y^2=\frac{9}{2}}\)

\(\orbr{\begin{cases}x=\frac{4}{z}\\y=\frac{6}{z}\end{cases}\Rightarrow\frac{24}{z^2}=3\Rightarrow z^2=8}\)

\(A=\frac{1}{2}\left(2+\frac{9}{2}+8\right)=\frac{4+9+16}{4}=\frac{29}{4}\)