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\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (DK \(x\ge1\)
\(=\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|}\)
vs \(x\ge2\) \(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\) \(\Rightarrow A=\sqrt{2x-2}\)
vs \(1\le x< 2\) \(\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2x-1}+1-1+\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\) \(\Rightarrow A=\frac{\sqrt{2}}{\sqrt{2x-1}}\)
\(\sqrt{2X-1}\ge1\Leftrightarrow X\ge1\)NEN SUY RA THEO CACH LAM CUA TO
THOI U AM BUSY SEE YOU AGAIN
\(A=\left(a-1\right)\sqrt{\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-a\sqrt{\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)^2.\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-\sqrt{a^2.\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)
\(A=\sqrt{a\left(a-1\right)}\)
Bài 2:
a, Ta có
\(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}\)
= \(3\left|-2\right|+\left|-5\right|\)
=\(6+5\)
= 11
Vậy \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}=11\)
b, Ta có
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}\)
= \(\left|\sqrt{5}+1\right|-\sqrt{5}\)
= \(\sqrt{5}+1-\sqrt{5}=1\)
Vậy \(\sqrt{6+2\sqrt{5}}-\sqrt{5}=1\)