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A2=x+\(\sqrt{2x-1}\)+x-\(\sqrt{2x-1}\)- 2\(\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}\)
A2=2x-2\(\sqrt{x^2-2x+1}\)
A2=2x-2(x-1)=1
=>A=1(vì a>0)
Ta có: \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\) \(\left(ĐK:x\ge\frac{1}{2}\right)\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1}+1-\sqrt{2x-1}+1\)
\(\Leftrightarrow A\sqrt{2}=2\)
\(\Leftrightarrow A=\sqrt{2}\)
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\(A=\frac{\sqrt{x}-1}{x^2-x}:\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right)\)
\(A=\frac{\sqrt{x}-1}{x\left(x-1\right)}:\left(\frac{\sqrt{x}+1-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\frac{1}{x\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\frac{1}{x\left(\sqrt{x}+1\right)}.\left(\sqrt{x}+1\right)\)
\(A=\frac{1}{x}\)
A = x 2 − x x − 1 : x 1 − x + 1 1 A = x x − 1 x − 1 : x x + 1 x + 1 − 1 A = x x + 1 1 : x x + 1 x A = x x + 1 1 . x + 1 A = x 1 √ ( √ √ ) √ ( ) ( √ √ (√ ) ) (√ ) √ √ (√ ) (√ ) (√ )
A = x 2 − x x − 1 : x 1 − x + 1 1 A = x x − 1 x − 1 : x x + 1 x + 1 − 1 A = x x + 1 1 : x x + 1 x A = x x + 1 1 . x + 1 A = x 1 √ ( √ √ ) √ ( ) ( √ √ (√ ) ) (√ ) √ √ (√ ) (√ ) (√ )\(A = x 2 − x x − 1 : x 1 − x + 1 1 A = x x − 1 x − 1 : x x + 1 x + 1 − 1 A = x x + 1 1 : x x + 1 x A = x x + 1 1 . x + 1 A = x 1 √ ( √ √ ) √ ( ) ( √ √ (√ ) ) (√ ) √ √ (√ ) (√ ) (√ )\)
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A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
Ta có:
A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
= \(\dfrac{-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-1\)
Kết luận: ...
ĐK của nó còn là: x ≥ 0 nữa dung doan nhé, mình viết thiếu...
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\(A=\sqrt{2}-\sqrt{x+2\sqrt{2x-4}}\) ( ĐKXĐ: \(x\ge2\))
\(\Rightarrow A\sqrt{2}=2-\sqrt{2x+4\sqrt{2x-4}}\)
\(=2-\sqrt{\left(\sqrt{2x-4}+2\right)^2}\)
\(=2-\sqrt{2x-4}-2\)
\(=-\sqrt{2x-4}\)
\(\Rightarrow A=-\sqrt{\frac{2x-4}{2}}\)
\(=-\sqrt{x-2}\)
\(A=-1\Leftrightarrow-\sqrt{x-2}=-1\)
\(\Leftrightarrow\sqrt{x-2}=1\)
\(\Leftrightarrow x=3\)( Thỏa mãn ĐKXĐ )
TK NHA!
đkxđ x >=-1/2
\(\sqrt{2}A=\sqrt{2x-1+2\sqrt{2x-1}+1}.\sqrt{2x-1-2\sqrt{2x-1}+1}\)
đề có sai ko vậy