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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: \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\) \(\left(ĐK:x\ge\frac{1}{2}\right)\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1}+1-\sqrt{2x-1}+1\)
\(\Leftrightarrow A\sqrt{2}=2\)
\(\Leftrightarrow A=\sqrt{2}\)
\(A=\frac{\sqrt{x}-1}{x^2-x}:\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right)\)
\(A=\frac{\sqrt{x}-1}{x\left(x-1\right)}:\left(\frac{\sqrt{x}+1-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\frac{1}{x\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\frac{1}{x\left(\sqrt{x}+1\right)}.\left(\sqrt{x}+1\right)\)
\(A=\frac{1}{x}\)
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 √ ( √ √ ) √ ( ) ( √ √ (√ ) ) (√ ) √ √ (√ ) (√ ) (√ )\(A = x 2 − x x − 1 : x 1 − x + 1 1 A = x x − 1 x − 1 : x x + 1 x + 1 − 1 A = x x + 1 1 : x x + 1 x A = x x + 1 1 . x + 1 A = x 1 √ ( √ √ ) √ ( ) ( √ √ (√ ) ) (√ ) √ √ (√ ) (√ ) (√ )\)
\(A=\sqrt{2}-\sqrt{x+2\sqrt{2x-4}}\) ( 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!
Bài 2:
a: \(A=2\sqrt{7}-1+\left(\sqrt{7}+4\right)\)
\(=2\sqrt{7}-1+\sqrt{7}+4=3\sqrt{7}+3\)
b: \(B=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
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