Phân tích đa thức sau thành nhân tử
X2-xz+3yz-9y2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,=4x^2-1\\ 2,=\left(x-4\right)^2-9y^2=\left(x-3y-4\right)\left(x+3y-4\right)\)
1)\(\left(2x+1\right)\left(2x-1\right)=\left(2x\right)^2-1^2=4x^2-1\)
2)\(x^2-8x-9y^2+16=\left(x^2-8x+16\right)-9y^2=\left(x^2-8x+4^2\right)-\left(3y\right)^2=\left(x-4\right)^2-\left(3y\right)^2=\left[\left(x-4\right)-3y\right]\left[\left(x-4\right)+3y\right]=\left(x-4-3y\right)\left(x-4+3y\right)\)
a) x2-xz-9y2+3yz
=(x2-9y2)-(xz-3yz)
=(x-3y)(x+3y)-z(x-3y)
=(x-3y)(x+3y-z)
b)x3-x2-5x+125
=x3-6x2+25x+5x2-30x+125
=x(x2-6x+25)+5(x2-6x+25)
=(x+5)(x2-6x+25)
c.x3+2x2-6x-27
=x3+5x2+9x-3x2-15x-27
=x(x2+5x+9)-3(x2+5x+9)
=(x-3)(x2+5x+9)
d. 12x3+4x2-27x-9
=12x3+4x2-27x-9
=4x2(3x+1)-9(3x+1)
=(4x2-9)(3x+1)
=(2x-3)(2x+3)(3x+1)
e.x4-25x2+20x-4
=x4+5x3-2x2-5x2-25x+10+2x2+10x-4
=x2(x2+5x-2)-5(x2+5x-2)+2(x2+5x-2)
=(x2-5x+2)(x2+5x-2)
f.x2(x2-6)-x2+9
=x4+x3-3x2-x3-x2+3x-3x2-3x+9
=x2(x2+x-3)-x(x2+x-3)-3(x2+x-3)
=(x2-x-3)(x2+x-3)
\(x^2-xz-9y^2+3yz\)
\(=\)\(\left(x^2-9y^2\right)-\left(xz-3yz\right)\)
\(=\)\(\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\)\(\left(x-3y\right)\left(x+3y-z\right)\)
Chúc bạn học tốt ~
\(x^2-xz-9y^2+3yz\)
\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
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`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
\(x^2-11x+3\\ =\left(x^2-4x+4\right)-7x-1\\ =\left(x-2\right)^2-\left(\sqrt{7x+1}\right)^2\\ =\left(x-2-\sqrt{7x+1}\right)\left(x-2+\sqrt{7x+1}\right)\)
\(=x^2+x-6x+6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x+6\right)\)
\(x^2-3x-4=x^2+x-4x-4=x\left(x+1\right)-4\left(x+1\right)=\left(x-4\right)\left(x+1\right)\)
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. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$