Phân tích đa thức thành nhân tử

c: \(x^2-4x^2y^2+y^2+2xy\)

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

\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)

d: \(7x^2-14xy^2+7y^4\)

\(=7\left(x^2-2xy^2+y^4\right)\)

\(=7\left(x-y^2\right)^2\)

\(=2y^2+3xy-x^2\)

\(14x^2-14xy-8x+8y=14x\left(x-y\right)-8\left(x-y\right)=\left(x-y\right)\left(14x-8\right)\)

\(14x^2-14xy-8x+8y\)

\(=14x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(14x-8\right)\left(x-y\right)\)

\(=2\left(7x-4\right)\left(x-y\right)\)

\(14x^2-14xy-8y+8y=2\left(x-y\right)\left(7x-4\right)\)

\(8x^2+14xy+8y^2+2x-2y+2=0\)

\(\Leftrightarrow7\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow7\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

Do \(\left\{{}\begin{matrix}7\left(x+y\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) ; \(\forall x;y\)

Nên \(7\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0;\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

7x²-14xy²+7y⁴=7(x²-2xy²+(y²)²) =7(x-y²)²

Nếu câu hỏi là phân tích đa thức thành nhân tử:

\(7x^2-14xy^2-7y^4=7\left(x^2-2xy^2-\left(y^2\right)^2\right)=7\left(x-y^2\right)^2\)

Phan h da thuc thanh nhan tu a ban ?

Ta co 

7x²-14xy+7y²

=7(x²-2xy+y²)

=7(x-y)²

Ta có:\(7x^2-14xy+7y^2\)

\(=7\left(x^2-2xy+y^2\right)\)

\(=7\left(x^2-2.x.y+y^2\right)\)

\(=7\left(x-y\right)^2\)

a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)

b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

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

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

c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)

d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)

5x . (3x + 2y) + 7x . (8y - 4x)

= 15x^2 + 10xy + 56xy - 28x^2

= 66xy - 13x^2