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Giúp mk với 

Ý bạn là như này phải ko ?

 \(\frac{4}{15x^3.y^5}\)

\(\frac{11}{12x^4.y^2}\)

Ta có:

\(\frac{x}{x^2+x+1}=-\frac{1}{4}\Rightarrow x^2+x+1=-4x\)

\(\Rightarrow x^2+5x+1=0\Rightarrow x^2=5x+1\)

Với x²=5x+1 ta được:

\(P=\frac{2x\left(5x+1\right)^2+10\left(5x+1\right)^2+2x\left(5x+1\right)-7\left(5x+1\right)-35x+2009}{2029+60x+11\left(5x+1\right)-5x\left(5x+1\right)-\left(5x+1\right)^2}\)

\(P=\frac{2x\left(25x^2+10x+1\right)+10\left(25x^2+10x+1\right)+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-\left(25x^2+10x+1\right)}\)

\(P=\frac{50x^3+20x^2+2x+250x^2+100x+10+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-25x^2-10x-1}\)

\(P=\frac{50x^3+280x^2+34x+2012}{2039+100x-50x^2}\)

\(P=\frac{50x\left(5x+1\right)+280\left(5x+1\right)+34x+2012}{2039+100x-50\left(5x+1\right)}\)

\(P=\frac{250x^2+50x+1400x+280+34x+2012}{2039+100x-250x-50}\)

\(P=\frac{250\left(5x+1\right)+50x+1400x+280+34x+2012}{1989-150x}\)

\(P=\frac{1250x+250+50x+1400x+280+34x+2012}{1989-150x}\)

bài trên sai rồi

THAY 2018 = xyz vào biểu thức 

      \(\frac{xyzx}{xy+xyzx+xyz}\)  +  \(\frac{y}{yz+y+xyz}\)+  \(\frac{z}{xz+z+1}\)

 =  \(\frac{xz}{1+xz+z}\)+  \(\frac{1}{z+1+xz}\)+  \(\frac{z}{xz+z+1}\)=  \(\frac{xz+z+1}{xz+z+1}\)=\(1\)

Đặt \(A=\frac{2018x}{xy+2018x+2018}+\frac{y}{yzz+y+2018}+\frac{z}{xz+z+1}\)

Thay \(xyz=2018\)vào A ta được 

\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

   \(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{1}{xz+z+1}\)

  \(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

  \(=\frac{xz+1+z}{xz+z+1}=1\)