phân tích thành nhân tử x-2 \(\sqrt{x}\) -3
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\(x+3\sqrt{x+2}-2\)
\(=x+2+3\sqrt{x+2}-4\)
\(=\left(\sqrt{x+2}\right)^2+4\sqrt{x+2}-\sqrt{x+2}-4\)
\(=\left(\sqrt{x+2}+4\right)\left(\sqrt{x+2}-1\right)\)
\(x+3\sqrt{x+2}-2=x+2+3\sqrt{x+2}-4\)
\(=x+2-\sqrt{x+2}+4\sqrt{x+2}-4\)
\(=\sqrt{x+2}\left(\sqrt{x+2}-1\right)+4\left(\sqrt{x+2}-1\right)\)
\(=\left(\sqrt{x+2}-1\right)\left(\sqrt{x+2}+4\right)\)
\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=2+\sqrt{3}+\sqrt{6}+2\sqrt{2}\)
\(=2+\sqrt{3}+\sqrt{2}\left(2+\sqrt{3}\right)=\left(2+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)
\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)\)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
ai giải hộ mình với
Đặt \(t=\sqrt{x}\Rightarrow t^2=x\)
Ta có: \(t^2-2t-3=t^2+t-3t-3=t\left(t+1\right)-3\left(t+1\right)=\left(t+1\right)\left(t-3\right)\)
hay: \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)