c: =5a(a-b)+7(a-b)

=(a-b)(5a+7)

a: \(=a\left(y^2-2yz+z^2\right)\)

\(=a\left(y-z\right)^2\)

b: \(=\left(x^2+6xy+9y^2\right)-16\)

=(x+3y)^2-16

=(x+3y+4)(x+3y-4)

c: \(=7\left(a-b\right)+\left(a-b\right)\left(a+b\right)\)

=(a-b)(7+a+b)

d: \(36x^4-13x^2\)

=x^2*36x^2-x^2*13

=x^2(36x^2-13)

f: x^2-2xy+y^2-49

=(x-y)^2-49

=(x-y-7)(x-y+7)

e: 2x^3-18x

=2x(x^2-9)

=2x(x-3)(x+3)

g: 2x+2y-x^2-xy

=2(x+y)-x(x+y)

=(x+y)(2-x)

h: (x^2+3)^2+16

=x^4+6x^2+25

=x^4+10x^2+25-4x^2

=(x^2+5)^2-4x^2

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

\(7a^2-7ax-9a+9x=\left(7a^2-7ax\right)-\left(9a-9x\right)=7a\left(a-x\right)-9\left(a-x\right)=\left(a-x\right)\left(7a-9\right)\)

7a(a-x)-9(a-x)

(7a-9)(a-x)

\(7a^2-7ax-9a+9x=7a\left(a-x\right)-9\left(a-x\right)=\left(7a-9\right)\left(a-x\right)\)

\(a^3+4a^2-7a-10\)

\(=\left(a^3+5a^2\right)-\left(a^2+5a\right)-\left(2a+10\right)\)

\(=a^2\left(a+5\right)-a\left(a+5\right)-2\left(a+5\right)\)

\(=\left(a^2-a-2\right)\left(a+5\right)\)

\(=\left(a^2-2a+a-2\right)\left(a+5\right)\)

\(=\left[a\left(a-2\right)+\left(a-2\right)\right]\left(a+5\right)\)

\(=\left(a+1\right)\left(a-2\right)\left(a+5\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

\(\left(2a-b-2\right)\left(a+3b+1\right)\)

Bạn ơi làm thế nào để ra kết quả đấy thế?

\(a^3+4a^2-7a-10\)

\(=a^3+3a^2+a^2-10a+3a-10\)

\(=\left(a^3+a^2\right)+\left(3a^2+3a\right)-\left(10a+10\right)\)

\(=a^2\left(a+1\right)+3a\left(a+1\right)-10\left(a+1\right)\)

\(=\left(a+1\right)\left(a^2+3a-10\right)\)

\(=\left(a+1\right)\left[\left(a^2+5a-2a-10\right)\right]\)

\(=\left(a+1\right)\left[a\left(a+5\right)-2\left(a+5\right)\right]\)

\(=\left(a+1\right)\left(a+5\right)\left(a-2\right)\)

a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc

= 2( a^3 + b^3 ) + 7ab(a+b) = 2(a+b)(a^2 -ab +b^2)  + 7ab(a+b)  = (a+b) ( 2a^2 - 2ab + 2b^2 - 7ab)

=(a +b ) ( 2a^2 +2b^2 - 9ab)

saj ruj thang Tran saj