\(x^2-4+5\left(4x-8\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+20\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2+20\right)=0\Leftrightarrow\left(x-2\right)\left(x+22\right)=0\Leftrightarrow x=2;x=-22\)

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)

\(\Leftrightarrow A=\left(x+y\right)\left(x^2+2xy+y^2\right)=\left(x+y\right)\left(x+y\right)^2=\left(x+y\right)^3\)

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)

\(\Leftrightarrow A=\left(x^2+y^2\right)\left(x+y\right)+2xy\left(x+y\right)\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)\left(x+y\right)\)

\(\Leftrightarrow A=\left(x+y\right)^2\left(x+y\right)\)

\(\Leftrightarrow A=\left(x+y\right)^3\)

\(\left(x+3\right)2-4\left(4-x\right)\left(4+x\right)=10\)

\(\Leftrightarrow2x+6-\left(16-x^2\right)=10\)

\(\Leftrightarrow x^2+2x-10-10=0\Leftrightarrow x^2+2x-20=0\)

Giải trên máy tính ta được:\(x_1=-1+\sqrt{21};x_2=-1-\sqrt{21}\)

ik

Để E > 1 

Thì \(\frac{x^2}{x-1}\)>1

<==>\(x-1\)>0

<==>x > 1

Vậy x > 1 thì E > 1

Mk ms hk lp 7

I don't know English very much so i can't answere your question. Sory about that :(

=3x^2   -15xy  - 3y^2  +  15xy - 1 -3x^2 + 3y^2 =-1

vậy biểu thức không phụ thuộc vào biến