Phân tích đa thức thành nhân tử:
M= 7\(\sqrt{x-1}\) - \(\sqrt{x^3-x^2}\)+ x -1 với x\(\ge\)1
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\(x+2\sqrt{x-1}=\left(x-1\right)+2\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)
\(x-4\sqrt{x-2}+2=\left(x-2\right)-4\sqrt{x-2}+4=\left(\sqrt{x-2}-2\right)^2\)
\(x+2\sqrt{x-1}=\left(\sqrt{x-1}+1\right)^2\)
\(x-4\sqrt{x-2}+2=\left(\sqrt{x-2}+4\right)^2\)
\(M=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}+\left(\sqrt{x-1}\right)^2=\sqrt{x-1}\left(7-x+\sqrt{x-1}\right)\)
\(=\sqrt{x-1}\left(6-\left(x-1\right)+\sqrt{x-1}\right)\)( đến đây bạn có thể đặt \(\sqrt{x-1}=t\),t>=0 rồi giải)
\(=-\sqrt{x-1}\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+2\right)\)
1: \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
2: \(x-16=\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)\)
3: \(9x-1=\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)\)
4: \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(1,x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\\ 2,x-16=\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)\\ 3,9x-1=\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)\\ 4,x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)
b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)
c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
Ta có : \(M=7\sqrt{x-1}-\sqrt{x^3-x^2}+x-1\)
\(=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}+x-1\)
\(=7\sqrt{x-1}-x\sqrt{x-1}+\left(\sqrt{x-1}\right)^2\)
\(=\sqrt{x-1}\left(7-x+\sqrt{x-1}\right)\)
\(=\sqrt{x-1}\left(\sqrt{x-1}+2\right)\left(\sqrt{x-1}-3\right)\)
CẢM ƠN BẠN