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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
= y^2-y+x^2-3x+2
= (x^2-3x+2,25)-(y^2+y+0,25)
= (x-3/2)^2 - (y+1/2)^2
= (x-3/2+y+1/2).(x-3.2-y-1/2)
=(x+y-1).(x-y-2)
k mk nha
x2 - x - y2 - y
= (x2 - y2) - (x + y)
= (x - y).(x + y) - (x + y)
= (x + y).(x - y - 1)
\(x^2-x-y^2-y\)
\(=x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x-y-1\right)\left(x+y\right)\)
y^2 - ( x^2 - 2x + 1 )
⇔ y^2 - ( x - 1 )^2
⇔ ( y - x - 1 ) ( y + x - 1 )
nha bạn
( x + y )2 - 2( x + y ) + 1 =
= ( x + y - 1 )2 ( áp dụng hằng đẳng thức thứ bình phương 1 hiệu nha )
Hok tốt!!!!!!!!!!!!!!!!
x2-x-y2-y
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\cdot\left(x-y-1\right)\)
********nha
\(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)\(=\left(x+y\right)\left(x-y-1\right).\)
\(\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y-1\right)^2\) (HĐT số 2)