Phân tích thành nhân tử
1/ x^5+x^4-x^3+x^2-x+2
2/ x^4+ 5x^3+ 10x- 4
ai giải giúp tớ đi
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NV
1
13 tháng 7 2015
x^4 + 5x^3 +10x -4
<=> (x^4 - 4)+ (5x^3 +10x)
<=> (x^2 - 2)(x^2 +2) + 5x(x^2 + 2)
<=> (x^2 + 2)(x^2 -2 + 5x)
Mk làm như z đó pn xem thử ^^
TP
3
21 tháng 8 2021
\(3x^2+x-2=3x^2-2x+3x-2=x\left(3x-2\right)+\left(3x-2\right)=\left(x+1\right)\left(3x-2\right)\)
\(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(x^2+2xy-15y^2=x^2-3xy+5xy-15y^2=x\left(x-3y\right)+5y\left(x-3y\right)=\left(x+5y\right)\left(x-3y\right)\)
NV
0
5 tháng 9 2017
a) \(5x^2\)\(\left(x-2y\right)\)\(-\)\(15x\)\(\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x^2-15x\right)\)
\(=5x\left(x-2y\right)\left(x-3\right)\)
b) \(3\left(x-y\right)\)\(-\)\(5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
c) \(10x\left(x-y\right)\)\(-\)\(8y\left(y-x\right)\)
\(=\)\(10x\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(x-y\right)\left(10x+8y\right)\)
\(=2\left(5x+4\right)\left(x-y\right)\)
d) \(x^2\)\(\left(x-5\right)\)\(+\)\(4\)\(\left(5-x\right)\)
\(=x^2\)\(\left(x-5\right)\)\(-\)\(4\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-4\right)\)
\(=\left(x-5\right)\left(x-2\right)\left(x-2\right)\)
5 tháng 9 2017
a) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x^2-15x\right)\)
\(=\left(x-2y\right)\left(x-3\right)5x\)
b)\(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(3+5x\right)\left(x-y\right)\)
c)\(10x\left(x-y\right)-8y\left(y-x\right)\)
\(=10x\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(10x+8y\right)\left(x-y\right)\)
\(=2\left(5x+4y\right)\left(x-y\right)\)
d)\(x^2\left(x-5\right)+4\left(5-x\right)\)
\(=x^2\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x^2-4\right)\left(x-5\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)
24 tháng 9 2019
a) \(x^4+5x^3+10x-4\)
\(=\left(x^4+2x^2\right)+\left(5x^3+10x\right)-\left(2x^2+4\right)\)
\(=x^2\left(x^2+2\right)+5x\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-2\right)\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\right]\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2\right]\)
\(=\left(x^2+2\right)\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x^2+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
b) \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-zx-zy+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)
NL
2
21 tháng 8 2021
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
TP
2
21 tháng 8 2021
1) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
21 tháng 8 2021
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
PA
1
AH
Akai Haruma
Giáo viên
4 tháng 10 2021
Lời giải:
a.
$x^4+10x^3+26x^2+10x+1$
$=(x^4+10x^3+25x^2)+x^2+10x+1$
$=(x^2+5x)^2+2(x^2+5x)+1-x^2$
$=(x^2+5x+1)^2-x^2=(x^2+5x+1-x)(x^2+5x+1+x)$
$=(x^2+4x+1)(x^2+6x+1)$
b.
$x^4+x^3-4x^2+x+1$
$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$
$=x^2(x-1)(x+1)+x^2(x-1)-x(x-1)-(x-1)(x+1)$
$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$
$=(x-1)(x^3+2x^2-2x-1)$
$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$
$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$
6 tháng 10 2021
a) \(-10x^3+2x^2=0\)
\(\Rightarrow-2x^2\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(5x\left(x-2016\right)-x+2016=0\)
\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)
\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)
6 tháng 10 2021
a: Ta có: \(-10x^3+2x^2=0\)
\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
thtrhtrh