Phân tích đa thức sau thành nhân tử:
x^2 - 5x + 6
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\(x^4-5x^2y^2+4y^4\)
\(=\left(x^2\right)^2-2x^22y^2+\left(2y^2\right)^2-x^2y^2\)
\(=\left(x^2-2y^2\right)^2-\left(xy\right)^2\)
\(=\left(x^2-2y^2-xy\right)\left(x^2-2y^2+xy\right)\)
Sửa đề: \(x^2+5xy+2xy+10y^2\)
=x(x+5y)+2y(x+5y)
=(x+5y)(x+2y)
\(x^2+5x-2=\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}-2=\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2=\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
\(=\left(x+\frac{5-\sqrt{33}}{2}\right)\left(x+\frac{5+\sqrt{33}}{2}\right)\)
Sửa đề: \(x^2+3xy+2xy+6y^2\)
\(=\left(x^2+3xy\right)+\left(2xy+6y^2\right)\)
=x(x+3y)+2y(x+3y)
=(x+3y)(x+2y)
\(=x^2+2x\cdot\frac{1}{2}+\frac{1}{4}-\left(\frac{\sqrt{23}}{2}i\right)^2\)
\(=\left(x+\frac{1}{2}\right)^2\)\(-\left(\frac{\sqrt{23}}{2}i\right)^2\)
\(\left(x+\frac{1}{2}-\frac{\sqrt{23}}{2}i\right)\left(x+\frac{1}{2}+\frac{\sqrt[]{23}}{2}i\right)\)
\(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(\dfrac{1}{4}x^2+2xy+4y^2=\left(\dfrac{1}{2}x+2y\right)^2\)
x2 - 5x + 6
= x2 - 2x - 3x + 6
= ( x2 - 2x ) - ( 3x - 6 )
= x( x - 2 ) - 3( x - 2 )
= ( x - 3 )( x - 2 )
Tìm x biết:
(x+1)^2 + 2x (x-2) = 3 (x+4) (x+1)