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\(A=\sqrt{\left(\sqrt{a-1}+1\right)^2}+\sqrt{\left(\sqrt{a-1}-1\right)^2}\)
\(A=\sqrt{a-1}+1+1-\sqrt{a-1}\) ( DO: a < 2 - gt => \(1>\sqrt{a-1}\))
\(A=2\)
Vậy A = 2.
a) ĐKXĐ: \(\hept{\begin{cases}2x-1\ge0\\2x\ge2\sqrt{2x-1}\end{cases}}\)\(\Leftrightarrow x\ge\frac{1}{2}\)
A=\(\sqrt{2x-1+1+2\sqrt{2x-1}}\)\(-\sqrt{2x-1+1-2\sqrt{2x-1}}\)
=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}\)\(-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
=\(\sqrt{2x-1}+1-|\sqrt{2x-1}-1|\)
Nếu \(x\ge1\)thì A=\(\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)\)=2.
Nếu \(\frac{1}{2}\le x< 1\)thì A=\(\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)\)=\(2\sqrt{2x-1}\).
b)A<1 thì \(\frac{1}{2}\le x< 1\)và \(2\sqrt{2x-1}< 1\)\(\Leftrightarrow4\left(2x-1\right)< 1\)\(\Leftrightarrow8x-4< 1\)\(\Leftrightarrow x< \frac{5}{8}\)(tm)
Vậy A<1 thì \(\frac{1}{2}\le x< \frac{5}{8}\).
Bài làm:
\(\frac{1}{x-1}.\sqrt{x^2-2x+1}\)
\(=\frac{1}{x-1}.\sqrt{\left(x-1\right)^2}\)
\(=\frac{1}{x-1}.\left|x-1\right|\)
\(=\frac{1}{x-1}.-\left(x-1\right)\)(Vì x < 1 )
\(=-1\)
\(B=\sqrt{x+\sqrt{x^2-1}}-\sqrt{x-\sqrt{x^2-1}}\)
\(B^2=x+\sqrt{x^2-1}+x-\sqrt{x^2-1}-2\sqrt{\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)}\)
\(B^2=2x-2\sqrt{x^2-x^2+1}\)
\(B^2=2x-2\)
\(\Rightarrow B=\sqrt{2x-2}\)
\(C=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-1}\left(ĐK:x\ge1\right)\)
\(C=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{x-1}\)
\(C=\sqrt{x-1}+1-\sqrt{x-1}=1\)
\(A=\left(a-1\right)\sqrt{\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-a\sqrt{\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)^2.\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-\sqrt{a^2.\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)
\(A=\sqrt{a\left(a-1\right)}\)
\(\sqrt{x^2-2x+1}+x\\ =\sqrt{\left(x-1\right)^2}+x\\ =\left|x-1\right|+x\\ =x-1+x\\ =2x-1\)
`->A`