(b-a-5)*(b-a)

a²+b²-5b-2ab+5a
= (b²-2ab+a²)-5(b-a)
= (b-a)²-5(b-a)
= (b-a)(b-a-5)

= (2a-b+1)(a+2b-3)

=(5a²+5b²+10ab)-5c²=5.(a²+2ab+b²)-5c²=5.(a+b)²-5c²=5.[(a+b)²-c²]=5.(a+b+c).(a+b-c)

=(5a²+5b²+10ab)-5c²=5.(a²+2ab+b²)-5c²=5.(a+b)²-5c²=5.[(a+b)²-c²]=5.(a+b+c).(a+b-c)

Bạn xem lại đề bài nhé.

ax²-5x²-ax+5x+a-5

=x^2(a-5)-x(a-5)+(a-5)

=(a-5)(x^2-x+1)

cậu ghi sai đề rồi phải là

3ax²+3bx²+ax+bx+5a+5b

=3x^2(a+b)+x(a+b)+5(a+b)

=(a+b)(3x^2+x+5)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

   3a\(x\)² + 3b\(x\)² + a\(x\) + b\(x\) + 5a + 5b 

= (3a\(x^2\) + 3b\(x^2\)) + (a\(x\) + b\(x\)) + (5a + 5b)

= 3\(x^2\)(a + b) + \(x\)(a +b) + 5(a + b)

= (a + b)( 3\(x^2\) + \(x\) + 5)

= (a + b)(3\(x^2\) + \(x\) + 5)²