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Bài 1:
a: \(49-4x^2=\left(7-2x\right)\left(7+2x\right)\)
b: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
c: \(x^2+18xy+81y^2=\left(x+9y\right)^2\)
Câu 1:
\(4x^2+16x-9\)
\(=4x^2+18x-2x-9\)
\(=2x\left(2x+9\right)-\left(2x+9\right)\)
\(=\left(2x-1\right)\left(2x+9\right)\)
Câu 2:
\(6x^2-11x+3=0\)
\(\Leftrightarrow6x^2-2x-9x+3=0\)
\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
a. `6x(x-2015)-x+2015=6x(x-2015)-(x-2015)=(x-2015)(6x-1)`
b. `x^4+4x^2+4=(x^2)^2+2.x^2 .2 +2^2=(x^2+2)^2`
a) \(6x\left(x-2015\right)-x+2015\)
\(=6x\left(x-2015\right)-\left(x-2015\right)\)
\(=\left(x-2015\right)\left(6x-1\right)\)
b) \(x^4+4x^2+4\)
\(=x^4+2\cdot x^2\cdot2+2^2\)
\(=\left(x^2+2\right)^2\)
Lời giải:
1.
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$
$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$
2.
$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$
3.
$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$
$=(x-1)(x^2+3x+1)$
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(x^2+y\right)\)
c) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
\(16x^3y+\frac{1}{4}yz^3\)
\(\text{Phân tích thành nhân tử}\)
\(\frac{y\left(\frac{z}{2}+2x\right)\left(z^2-4xz+16x^2\right)}{2}\)
\(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)
=x^3-3.x2.\(\frac{1}{4}\)+3.x.(\(\frac{1}{4}\))2-\(\frac{1}{4^3}\)+(3x)^3+3.(3x)2.\(\frac{1}{2}\)+3.3x.\(\frac{1}{2^2}\)+(\(\frac{1}{2}\))3
=(x-1/4)3+(3x+1/2)3