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 

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

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

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

  \(=\frac{xz+1+z}{xz+z+1}=1\)

Có: M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)

=> M = (a + b)(a² - ab + b²) + 3ab((a + b)² - 2ab) + 6a²b²(a + b)

=> M = (a + b)[(a + b)² - 3ab] + 3ab[(a + b)² - 2ab] + 6a²b²(a + b)

=> M = 1 - 3ab + 3ab(1 - 2ab) + 6a²b²     (vì a+b=1)

=> M = 1 - 3ab + 3ab - 6a²b² + 6a²b² 

=> M = 1

Vậy M = 1

M = \(a^3\)+ \(b^3\)+ 3ab ( \(a^2\)+ \(b^2\)) + \(6a^2\)\(b^2\)(a+b)

M = ( a + b ) ( \(a^2\)- ab + \(b^2\))  + 3ab [ \(a^2\)+ \(b^2\)+ 2ab( a + b )

M = \(a^2\)- ab + \(b^2\)+ 3ab ( \(a^2\)+ 2ab + \(b^2\))

Với a + b = 1

M= \(a^2\)- ab + \(b^2\)+ 3ab\(\left(a+b\right)^2\)

M = \(a^2\)- ab + \(b^2\)+ 3ab

M = \(a^2\)+ \(b^2\)+ 2ab

M = \(a^2\)+ 2ab + \(b^2\)

M = \(\left(a+b\right)^2\)

M = 1

Vậy M = 1

Èo,phân tích ra tưởng cái hệ 3 ẩn r định bỏ cuộc và cái kết:(

Ta có:

\(f\left(x\right)=\left(x-2\right)\cdot Q\left(x\right)+5\)

\(f\left(x\right)=\left(x+1\right)\cdot K\left(x\right)-4\)

Theo định lý Huy ĐZ ta có:

\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\left(1\right)\)

\(\Rightarrow f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\left(2\right)\)

Lấy \(\left(1\right)-\left(2\right)\) ta được:

\(9+3a+3b=9\Leftrightarrow a+b=0\)

Khi đó:

\(\left(a^3+b^3\right)\left(b^5+c^5\right)\left(c^7+d^7\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)\left(b^5+c^5\right)\left(c^7+a^7\right)\) 

\(=0\) ( theo Huy ĐZ thì \(a+b=0\) )

Ap dung dinh ly Bozout ta co

\(f\left(2\right)=2^3+a.2^2+b.2+c=5\)

<=> \(4a+2b+c=-3\) (1)

tuong tu \(f\left(-1\right)=\left(-1\right)^3+a-b+c=-4\)

<=> \(a-b+c=-3\) (2)

tu (1) va (2) => \(4a+2b=a-b=-3\) 

=> a=b+-3

=> \(4\left(b-3\right)+2b=-3\Rightarrow b=\frac{3}{2}\)

=> \(a=-\frac{3}{2}\)

=> \(\left(a^3+b^3\right)=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(\frac{3}{2}-\frac{3}{2}\right)\left(a^2-ab+b^2\right)=0\)

=> gia tri bieu thuc =0

\(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)

\(=\left(x^6:6x^2\right)+\left(-2x^5:6x^2\right)+\left(2x^4:6x^2\right)+\left(6x^3:6x^2\right)+\left(-4x^2:6x^2\right)\)

\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)