(2x^2+1/3)

= 2x^2+0.(3)

9x^2+4x+2-2y^2

=> -2y^2+9x^2+4x+2

=> -(2y^2-9x^2-4x-2)

nha bạn 

Phan tích đa thức thành nhân tử ạ

ta có :

\(\left(a^3-27\right)+\left(3-a\right)\left(4a+21\right)=\left(a-3\right)\left(a^2+3a+9-4a-21\right)\)

\(=\left(a-3\right)\left(a^2-a-12\right)=\left(a-3\right)\left(a^2-4a+3a-12\right)\)

\(=\left(a-3\right)\left(a+3\right)\left(a-4\right)\)

bài này là tìm x bạn ơi

\(x^2-4x+4+\left(x-2\right).\left(x+1\right)=0\)

\(\Rightarrow\left(x-2\right)^2+\left(x-2\right).\left(x+1\right)=0\)

\(\Rightarrow\left(x-2\right).\left(2x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\2x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)

Vậy .................

Chúc bn học tốt

\(4x^2+y^2-4x+2y+2=0\)

\(\Leftrightarrow4x^2-4x+1+y^2+2y+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y+1\right)^2=0\)

Đẳng thức xảy ra khi x = 1/2 ; y = - 1

Trả lời:

\(4x^2+y^2-4x+2y+2=0\)

\(\Leftrightarrow4x^2+y^2-4x+2y+1+1=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}}\)

Vậy x = 1/2; x = - 1 là nghiệm của pt.

\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+2b+b^2=2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+b^2-2ab=0\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a=b\).