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=(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)
\(5a^2-5b^2-20a+20b\)
\(=5\left(a^2-b^2\right)-20\left(a-b\right)\)
\(=5\left(a-b\right)\left(a+b\right)-20\left(a-b\right)\)
\(=\left[5\left(a+b\right)-20\right]\left(a-b\right)\)
\(=\left(5a+5b-20\right)\left(a-b\right)\)
\(5a^2-5b^2-20a+20=-5.\left(b-a+2\right).\left(b+a-2\right)\)
3a2- 10ab +3b2 =(3a2 -9ab) -( ab-3b2) = 3a(a-3b) - b(a-3b) =(a-3b)(3a-b)
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)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)