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\(\frac{1}{\sqrt{2}}.A=\frac{\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}}{\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+1}-\sqrt{\left(2x-1\right)-2\sqrt{2x-1}+1}}\)
\(=\frac{\sqrt{\left[\left(\sqrt{x-1}+1\right)\right]^2}+\sqrt{\left[\left(\sqrt{x-1}-1\right)^2\right]}}{\sqrt{\left[\sqrt{2x-1}+1\right]^2}-\sqrt{\left[\left(\sqrt{2x-1}\right)-1\right]^2}}\)
\(=\frac{\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|}{\left|\sqrt{2x-1}+1\right|-\left|\sqrt{2x-1}-1\right|}\)
DO X>2 NÊN TOÀN BỘ BIỂU THỨC TRONG TRỊ TUYỆT ĐỐI ĐỀU DƯƠNG
\(\frac{1}{\sqrt{2}}.A=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\)
=>\(A=\frac{\sqrt{x-1}}{\sqrt{2}}\)
\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (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
Q=\(\frac{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}{\sqrt{x+\sqrt{2x-1}-\sqrt{x-\sqrt{2x-1}}}}\)(x\(\ge2\))
\(=\frac{\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2}{\left(\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\right)\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)}\)
=\(\frac{2x+2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}}{2\sqrt{2x-1}}\)
=\(\frac{x+\sqrt{x^2-2x+1}}{\sqrt{2x-1}}=\frac{x+\sqrt{\left(x-1\right)^2}}{\sqrt{2x-1}}\) \(=\frac{x+x-1}{\sqrt{2x-1}}=\frac{2x-1}{\sqrt{2x-1}}=\sqrt{2x-1}\)
vậy \(Q=\sqrt{2x-1}với\)\(x\ge2\)
~ happy new year~
Happy new year !
Cố gắng kì sau làm CTV nha chị :D