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Xét : \(\sqrt{4x-3+4\sqrt{x-1}}=\sqrt{4\left(x-1\right)+4\sqrt{x-1}+1}=\sqrt{\left(2\sqrt{x-1}+1\right)^2}=2\sqrt{x-1}+1\)
Khi đó : \(A=\left(\sqrt{x-1}-1\right)^2+2\sqrt{x-1}=x-1-2\sqrt{x-1}+1+2\sqrt{x-1}+1=x+1\)
\(\sqrt{x+2\sqrt{x+1}}\)
\(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(\left|\sqrt{x-1}+1\right|\)
\(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}\right)^2-2\cdot\sqrt{x-1}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)
Gọi cái đó là A
A2 = x + \(2\sqrt{x-1}\) + x - \(2\sqrt{x-1}\)+
\(2\sqrt{\left(x+\sqrt{x-1}\right)\left(x-\sqrt{x-1}\right)}\)
= 2x + 2
Ta có:
\(A=x-\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{1}{\sqrt{x}+\sqrt{x-1}}\right)\)
\(A=x-\frac{\sqrt{x}+\sqrt{x-1}-\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}\)
\(A=x-\frac{2\sqrt{x-1}}{x-x+1}\)
\(A=x-2\sqrt{x-1}\)
\(A=\left(x-1\right)-2\sqrt{x-1}+1\)
\(A=\left(\sqrt{x-1}-1\right)^2\ge0\left(\forall x\ge1\right)\)
=> đpcm
\(P=\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(P=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(P=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1 \left(x\ge2\right)\) hoặc \(P=\sqrt{x-1}+1-\sqrt{x-1}+1\left(1\le x\le2\right)\)
\(\Rightarrow P=2\sqrt{x-1} \left(x\ge2\right)\) hoặc \(P=2 \left(1\le x\le2\right)\)