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\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
\(a,A=-3\sqrt{8}+\sqrt{50}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=-6\sqrt{2}+5\sqrt{2}+\left|1-\sqrt{2}\right|\)
\(=-\sqrt{2}-1+\sqrt{2}\)
\(=-1\)
Vậy \(A=-1\)
\(b,\)
\(=\left(\dfrac{5\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{5x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\dfrac{5\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{5\sqrt{x}-1}{\sqrt{x}}\)
Vậy \(B=\dfrac{5\sqrt{x}-1}{\sqrt{x}}\left(đk:x>0,x\ne1\right)\)
Sửa đề: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{9}{4}\end{matrix}\right.\)
a) Ta có: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\cdot\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2x+2\sqrt{x}+\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b) Ta có: \(x=\dfrac{3-2\sqrt{2}}{4}\)
\(\Leftrightarrow x=\dfrac{2-2\cdot\sqrt{2}\cdot1+1}{4}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\) vào biểu thức \(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\), ta được:
\(P=\left(3\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}-5\right):\left(2\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}+1\right)\)
\(\Leftrightarrow P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(\Leftrightarrow P=\left(\dfrac{3\cdot\left(\sqrt{2}-1\right)}{2}-\dfrac{10}{2}\right):\left(\sqrt{2}-1+1\right)\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-3-10}{2}:\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-13}{2}\cdot\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{6-13\sqrt{2}}{2}\)
Vậy: Khi \(x=\dfrac{3-2\sqrt{2}}{4}\) thì \(P=\dfrac{6-13\sqrt{2}}{2}\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
\(=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được:
\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\dfrac{1}{\sqrt{2}-1}\)
\(=\sqrt{2}+1\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
a) Ta có: \(B=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{3\sqrt{x}-6-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-8}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để \(B=\dfrac{1}{3}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}=2\)
\(\Leftrightarrow x=1\)(thỏa ĐK)
a) B= \(\left(\dfrac{3\left(\sqrt{x}-2\right)-1\left(\sqrt{x}+2\right)}{x-4}\right):\left(\dfrac{\sqrt{x}-6+1\left(\sqrt{x}-2\right)}{x-2\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{2\sqrt{x}-8}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để B=\(\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\sqrt{x}\Rightarrow x=1\)
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)