Phân tích đa thức đa thức thành nhân tử : 4x^2-y^2+4x+1
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\(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\\ ---\\ 4x^2-4x-3\\ =4x^2-4x+1-4\\ =\left(2x-1\right)^2-2^2=\left(2x-1-2\right)\left(2x-1+2\right)\\ =\left(2x-3\right)\left(2x+1\right)\)
1: =(2x)^2-2*2x*1+1^2
=(2x-1)^2
2: =4x^2-6x+2x-3
=2x(2x-3)+(2x-3)
=(2x-3)(2x+1)
\(4x^2+2xy+4x+y+1\)
\(=\left(4x^2+2x\right)+\left(2xy+y\right)+\left(2x+1\right)\)
\(=2x\left(2x+1\right)+y\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+y+1\right)\left(2x+1\right)\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)=\left(x-y\right)\left(x+y-4\right)\)
Sửa đề: 4x^2-y^2+8x+4
=4x^2+8x+4-y^2
=(2x+2)^2-y^2
=(2x+y+2)(2x-y+2)
4x2-y2+4x+1
=(4x2+4x+1)-y2
=(2x+1)2-y2
=(2x+1-y)(2x+1+y)
bài này tớ cũng ko chắc:
\(4x^2-y^2+4x+1=\left(4x+4x^2+1\right)-y^2= \left(2x+1\right)^2-y^2\)
\(=\left(2x+1\right)\left(2x+1\right)-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)