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1: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{2x-2\sqrt{2x-1}}-\sqrt{2x+2\sqrt{2x-1}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{2x-1}-1\right|-\left|\sqrt{2x-1}+1\right|\right)\)
TH1: x>=1
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{2x-1}-1-\sqrt{2x-1}-1\right)=-\sqrt{2}\)
TH2: 1/2<=x<1
\(A=\dfrac{1}{\sqrt{2}}\left(1-\sqrt{2x-1}-\sqrt{2x-1}-1\right)=-\sqrt{4x-2}\)
2:
\(=\sqrt{x-1+6\sqrt{x-1}+9}-\sqrt{x-2-2\sqrt{x-2}+1+3}\)
\(=\sqrt{x-1}+3-\sqrt{\left(\sqrt{x-2}-1\right)^2+3}\)
A = \(\frac{8\sqrt{41}}{2\sqrt{2^2+2.2.\sqrt{41}+\sqrt{41}^2}}\)
A = \(\frac{8\sqrt{41}}{2\sqrt{\left(2+\sqrt{41}\right)^2}}\)
A = \(\frac{8\sqrt{41}}{2\left|2+\sqrt{41}\right|}\)
A = \(\frac{8\sqrt{41}}{4+2\sqrt{41}}\)
B = \(\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1+x+4}{x+\sqrt{x}+1}\)
B = \(\left(\frac{2x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right).\frac{x+\sqrt{x}+1}{2x+\sqrt{x}+5}\)
Bạn tự làm tiếp nhé, mỏi tay quá!!
\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)
\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)
\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)
\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(\Leftrightarrow A^2=2x+2\sqrt{x^2-8x+16}=\)
\(=2x+\sqrt{\left(x-4\right)^2}\)
\(=2x+|x-4|\)
\(=\hept{\begin{cases}2x-x+4=x+4\left(2\le x< 4\right)\\2x+x-4=3x-4\left(x\ge4\right)\end{cases}}\)
\(\Rightarrow A=\hept{\begin{cases}\sqrt{x+4}\left(2\le x< 4\right)\\\sqrt{3x-4}\left(x\ge4\right)\end{cases}}\)
ĐKXĐ: \(x\ge2\)
\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{x-2}+\sqrt{2}\right|+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
Xét \(x\ge4\Rightarrow\sqrt{x-2}\ge\sqrt{2}\)
\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
Xét \(0\le x< 4\Rightarrow\sqrt{x-2}< \sqrt{2}\)
\(\Rightarrow A=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
Sửa đề: x-4
\(A=\dfrac{x-2\sqrt{x}+x+4\sqrt{x}+4+2x+8}{x-4}=\dfrac{4x+2\sqrt{x}+12}{x-4}\)
\(A=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)
\(B=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(=\sqrt{x-2+2\sqrt{2}\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}\sqrt{x-2}+2}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{2-x}\right)^2}\)
\(=\sqrt{2}+\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\)
\(A^2=x+2\sqrt{2x-4}+x-2\sqrt{2x-4}+2\sqrt{\left(x+2\sqrt{2x-4}\right)\left(x-2\sqrt{2x-4}\right)}=2x+2\sqrt{x^2-4\left(2x-4\right)}=2x+2\sqrt{x^2-8x+16}=2x+2\sqrt{\left(x-4\right)^2}=2x+2\left|x-4\right|\)
Suy ra A=$\sqrt{2x+2|x-4|}
Lời giải:
\(A\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}\)
\(=\sqrt{(2x-4)+4\sqrt{2x-4}+4}+\sqrt{(2x-4)-4\sqrt{2x-4}+4}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}+\sqrt{(\sqrt{2x-4}-2)^2}\)
\(=|\sqrt{2x-4}+2|+|\sqrt{2x-4}-2|\)
Nếu $x\geq 4$ thì:
$A=\sqrt{2x-4}+2+\sqrt{2x-4}-2=2\sqrt{2x-4}$
Nếu $2\leq x<4$ thì:
$A=\sqrt{2x-4}+2+2-\sqrt{2x-4}=4$