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a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-20\)
\(=\left(x^2+5x+4\right)\cdot\left(x^2+5x+6\right)-20\)
Đặt: \(x^2+5x+5=a\)Khi đó ta có:
\(A=\left(a-1\right)\left(a+1\right)-20=a^2-21=\left(a-\sqrt{21}\right)\left(a+\sqrt{21}\right)\)
tự thay trở lại
a) Đăt \(x^2+x=t\) khi đó bt trở thành:
\(t^2-2t-15=t^2+3t-5t-15=t\left(t+3\right)-5\left(t+3\right)\\ =\left(t+3\right)\left(1-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
câu 1:
x2y+xy2-x-y
=(x2y-x)+(xy2y)
=x(xy-1)+y(xy-1)
=(x+y)(xy-1)
Câu 2:sai đề
Câu 3:
ax2+ay-bx2-by
=(ax2+ay)-(bx2+by)
=a(x2+y)-b(x2+y)
=(a-b)(x2+y)
Câu 4:
x(x+1)2+x(x-5)-5(x+1)2
=(x+1)2(x-5)+x(x-5)
=(x+5)[(x+1)2+x]
=(x+5)(x2+3x+1)
Câu 5:
3x2-12y2
=3(x2-4y2)
=3[(x)2-(2y)2]
=3(x-2y)(x+2y)
Câu 6:
5xy2-10xyz+5xz2
=5x(y2-2yz+z2)
=5(z-y)2
a. 5x(x-3)(x+3)
b. (x-3)(x-4)
a. \(5x\left(x^2-9\right)\)
= \(5x\left(x+3\right)\left(x-3\right)\)
b.\(x^2-4x-3x+12\)
= \(\left(x^2-4x\right)-\left(3x-12\right)\)
\(=x\left(x-4\right)-3\left(x-4\right)\)
= \(\left(x-3\right)\left(x-4\right)\)