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a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)
\(=4x\left(a-b\right)-6xy\left(a-b\right)\)
\(=\left(4x-6xy\right)\left(a-b\right)\)
\(=2x\left(2-3y\right)\left(a-b\right)\)
a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)
b) \(=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
c) \(=x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-x+4\right)\)
Bài 3:
a: =>(2x-7)(x-2)=0
=>x=7/2 hoặc x=2
b: =>(x-1)(x+2)=0
=>x=1 hoặc x=-2
d: =>2x+3=0
hay x=-3/2
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
Đặt \(2x^2+x=t\)
Ta có: \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Rightarrow t^2-4t+3=0\)
\(\Leftrightarrow t=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot3}}{2\cdot1}\)
\(\Leftrightarrow t=\dfrac{4\pm\sqrt{16-12}}{2}\)
\(\Leftrightarrow t=\dfrac{4\pm\sqrt{4}}{2}\)
\(\Leftrightarrow t=\dfrac{4\pm2}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{4-2}{2}\\t=\dfrac{4+2}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+x=3\\2x^2+x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{3}{2};x_2=-1;x_3=\dfrac{1}{2};x_4=1\)
bài 2:
Đặt \(t=2x^2+x\) thì ta có:
\(t^2-4t+3=0\)\(\Rightarrow\left(t-3\right)\left(t-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t-1=0\\t-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x^2+x=1\\2x^2+x=3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x^2+x-1=0\\2x^2+x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)\left(2x-1\right)=0\\\left(x-1\right)\left(2x+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)