Ta có:\(x^2-4\)

\(=x^2-2^2\)

\(=\left(x-2\right)\left(x+2\right)\)

\(x^2-4\)

\(=x^2-2^2\)

\(=\left(x-2\right)\left(x+2\right)\)

Ta có : \(x^2-4\)

\(=x^2-2^2\)

\(=\left(x-2\right)\left(x+2\right)\)

\(x\left(x-y\right)+2\left(y-x\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)

\(=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)

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

\(=x\left(\frac{x^2}{4}+x+1\right)=x\left(\frac{x}{2}+1\right)^2\)

\(x^5-x^4-x^3-x^2-x-2=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)

\(=x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)

\(=\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)\)

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

\(x^2+7x+6\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right).\left(x+6\right)\)

\(x^2+7x+6=\left(x-6\right)\left(x-1\right)\)

-_-

Bài làm

x² + 7x + 6

<=> x² + x + 6x + 6

<=> x( x + 1 ) + 6( x + 1 )

<=> ( x + 1 )( x + 6 )

~ lớp 6 đã học đâu ~

\(x^2+3x+2\)

\(=x^2+x+2x+2\)

\(=x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(x+2\right)\)

(x+1)*(x+2)

1/3 số cam và 1/5 số quýt là 30 quả

1/4 số cam và 1/2 số quýt là 40 quả

Hỏi số cam và số quýt ?

giúp nhé

sai đề rồi bạn ơi

Bài nài cách làm đơn giản ạ :>>>

\(x^2+5x-6\)

\(=x^2-1x+6x-6\)

\(=\left(x^2-1x\right)+\left(6x-6\right)\)

\(=x\left(x-1\right)+6\left(x-1\right)\)

\(=\left(x-1\right)\left(x+6\right)\)

Hok tốt

x^2+2x+3x-6

=(x^2+2x)-(3x+6)

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

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

=(x-3)(x^2-4)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)