1.phân tích đa thức thành nhân tử
b) x2+y-xy-y
c) xy+z+y+xz
d) x2+(a+b)x+ab
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AH
Akai Haruma
Giáo viên
23 tháng 7 2021
Lời giải:
a. $a^4+a^3+a^2+a=(a^4+a^3)+(a^2+a)$
$=a^3(a+1)+a(a+1)=(a+1)(a^3+a)=a(a+1)(a^2+1)$
b. $3xy^2+5y-3x^2y+(-5x)=(3xy^2-3x^2y)+(5y-5x)$
$=3xy(y-x)+5(y-x)=(y-x)(3xy+5)$
c. $xy-z+y-xz=(xy+y)-(z+xz)=y(x+1)-z(x+1)=(x+1)(y-z)$
d.
$x^2-bx+ax-ab=(a^2+ax)-(bx+ab)=a(a+x)-b(a+x)=(a+x)(a-b)$
MH
25 tháng 10 2021
a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
25 tháng 10 2021
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
H
3
29 tháng 8 2021
\(xy-3x-2y+6=x\left(y-3\right)-2\left(y-3\right)=\left(y-3\right)\left(x-2\right)\)
\(x^2-6xy-4z^2+9y^2=\left(x-3y\right)^2-\left(2z\right)^2=\left(x-3y-2z\right)\left(x-3y+2z\right)\)
H
19 tháng 12 2023
Bài `1`
\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)
Bài `3`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
AH
Akai Haruma
Giáo viên
6 tháng 1
Lời giải:
a. $5x^2-10xy=5x(x-2y)$
b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$
d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$
e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$
f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$
4 tháng 8 2023
\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
4 tháng 8 2023
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
\(b,=x^2-xy=x\left(x-y\right)\\ c,=y\left(x+1\right)+z\left(x+1\right)=\left(y+z\right)\left(x+1\right)\\ d,=x^2+ax+bx+ab\\ =x\left(x+a\right)+b\left(x+a\right)=\left(x+b\right)\left(x+a\right)\)