\(a,A=5x+8xy+5y=(5x+5y)+8xy\)

\(=5(x+y)+8xy\)

\(=5\cdot\frac{2}{5}+8\cdot(-1)=2+(-8)=-6\)

\(b,B=2xy+7xyz-2xz\)

\(=2\cdot\frac{3}{7}y+7\cdot\frac{3}{7}yz-2\cdot\frac{3}{7}z\)

\(=\frac{6}{7}y+3yz-\frac{6}{7}z\)

\(=\frac{6}{7}y+3\cdot(-1)-\frac{6}{7}z\)

\(=\frac{6}{7}y+(-3)-\frac{6}{7}z\)

Làm nốt :v

a)

A=\(5\left(x+y\right)+8xy\) 

 \(=5.\frac{2}{5}+8.\left(-1\right)\) 

  \(=2-8\) 

   \(=-6\) 

Vậy.......

hc tốt

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

\(\Leftrightarrow\frac{3}{4}x-\frac{1}{2}=2\text{x}-8+\frac{1}{4}x\)

\(\Leftrightarrow\frac{3}{4}x-2\text{x}-\frac{1}{4}x=-8+\frac{1}{2}\)

\(\Leftrightarrow\frac{3-8-1}{4}x=\frac{-15}{2}\)

\(\Leftrightarrow-\frac{3}{2}x=-\frac{15}{2}\Leftrightarrow x=\frac{-15}{-3}=5\)

Vậy x = 5 

\(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-1\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-1\right)\cdot\frac{2}{9}=\frac{16}{9}\)

\(\Rightarrow\left(x-1\right)=\frac{16}{9}\div\frac{2}{9}\)

\(\Rightarrow\left(x-1\right)=\frac{16}{9}\cdot\frac{9}{2}\)

\(\Rightarrow x-1=8\Rightarrow x=9\)

Vậy x = 9 

\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(\Rightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(\Rightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(\Rightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

\(\Rightarrow\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\div2\)

\(\Rightarrow\frac{x}{x+1}=\frac{2004}{2005}\)

\(\Rightarrow2005\text{x}=2004\left(x+1\right)\)

\(\Rightarrow2005\text{x}=2004\text{x}+2004\)

\(\Rightarrow2005\text{x}-2004\text{x}=2004\)

\(\Rightarrow x=2004\)

Vậy x = 2004 

bad boy ơi giúp mình bài 2 đi 

a) \(\dfrac{2x^2-2xy}{x^2+x-xy-y}\) \(\left(x\ne y;x\ne-1\right)\)

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

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

\(=\dfrac{2x}{x+1}\)

b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

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

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

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

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