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a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x-\sqrt{x}-1}{x-2\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{x-5}{x-\sqrt{x}-2}\right)\)
\(=\dfrac{x-x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
b: \(=x-4\sqrt{x}+3\sqrt{x}-12=\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)\)
#)Thắc mắc ?
Bạn ơi ! chỗ kia là \(\sqrt{x}-7hay\sqrt{x+7}\)thế ???????????????
#)Giải :
\(5\sqrt{x-1}-\sqrt{x-7}=3x-4\)
ĐKXĐ : \(x\ge1\)
Đặt \(\hept{\begin{cases}\sqrt{x-1}=a\ge0\\\sqrt{x+7=b>0}\end{cases}\Rightarrow3x-4}=\frac{25a^2-b^2}{8}\)
Phương trình trở thành :
\(5a-b=\frac{25a^2-b^2}{8}\Leftrightarrow\left(5a-b\right)\left(5a+b\right)=8\left(5a-b\right)\)
\(\Leftrightarrow\orbr{\begin{cases}5a-b=0\\5a+b=8\end{cases}\Leftrightarrow\orbr{\begin{cases}5\sqrt{x-1}=\sqrt{x+7}\\5\sqrt{x-1}+\sqrt{x+7}=8\end{cases}}}\)
\(TH1:5\sqrt{x+1}=\sqrt{x+7}\Leftrightarrow25\left(x-1\right)=x+7\Rightarrow x=\frac{4}{3}\)
\(TH2:5\sqrt{x-1}+\sqrt{x+7}=8\)
\(\Leftrightarrow5\sqrt{x-1}-5+\sqrt{x+7}-3=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{x-1}+1}+\frac{x-2}{\sqrt{x-7}+3}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x-7}+3}\right)=0\)
\(\Rightarrow x=2\)
Bài 3
\(A=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=2\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)=2\cdot\sqrt{\left(\sqrt{5}+1\right)^2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=2\cdot4=8\left(đpcm\right)\)
\(B=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{2-\sqrt{3}}\right)=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\left(đpcm\right)\)
Bài 4
\(P=\frac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{10}\left(\sqrt{2}+1\right)-\sqrt{6}\left(\sqrt{2}+1\right)}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=2+\sqrt{2}\)
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+2\sqrt{2}+2+2}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)