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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. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
\(=-12x^2\left(y-x\right)+18x^3\left(y-x\right)\)
\(=-6x^2\left(y-x\right)\left(2-3x\right)\)
a ) \(x^2-12x+36=\left(x-6\right)^2\)
b ) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
c ) \(1-27x^3=\left(1-3x\right)\left(1+3x+9x^2\right)\)
d ) \(\left(x-y\right)-4y^2\) ( Câu này bó tay )
e ) \(y^3+3y^2+3y+1=\left(y+1\right)^3\)
f ) \(x^2+14x+49=\left(x+7\right)^2\)
Wish you study well !!
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`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
a) \(x^2-12x+36=\left(x-6\right)^2\)
b) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
c) \(1-27x^3=\left(1-3x\right)\left(1+3x+9x^2\right)\)
d) mk chỉnh lại đề chút xíu nhé. đúng thì bn tham khảo:
\(\left(x-y\right)^2-4y^2=\left(x-y-2y\right)\left(x-y+2y\right)=\left(x-3y\right)\left(x+y\right)\)
e) \(y^3+3y^2+3y+1=\left(y+1\right)^3\)
f) \(x^2+14x+49=\left(x+7\right)^2\)
Bạn cần làm gì với những đa thức này?
a) \(25x^2y-10xy+30xy^2=xy\left(25x+30y-10\right)\)
b) \(16x^2-20x=4x\left(4x-5\right)\)
c) \(18x^3+12x^2-30x=x\left(18x^2+12x-30\right)=x\left(x-1\right)\left(18x+9\right)=96x\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)
d) \(14x\left(x+y\right)-4y\left(x+y\right)=\left(14x-4y\right)\left(x+y\right)\)
e) \(9x^3y^2+12xy^2+15x^2y^2=3xy^2\left(3x^2+5x+4\right)\)
f) \(x\left(x-3\right)-y\left(3-x\right)=x\left(x-3\right)+y\left(x-3\right)=\left(x+y\right)\left(x-3\right)\)
g) \(5x\left(x+4\right)-2y\left(4+x\right)=\left(x+4\right)\left(5x-2y\right)\)