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1.
\(A=\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)
\(=\sqrt[3]{\left(\sqrt{3}+1\right)^3}-\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
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a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x-\sqrt{x}-1}{x-2\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{x-5}{x-\sqrt{x}-2}\right)\)
\(=\dfrac{x-x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
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b: \(=x-4\sqrt{x}+3\sqrt{x}-12=\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)\)
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#)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 :
\(5a-b=\frac{25a^2-b^2}{8}\Leftrightarrow\left(5a-b\right)\left(5a+b\right)=8\left(5a-b\right)\)
\(\Leftrightarrow\orbr{\begin{cases}5a-b=0\\5a+b=8\end{cases}\Leftrightarrow\orbr{\begin{cases}5\sqrt{x-1}=\sqrt{x+7}\\5\sqrt{x-1}+\sqrt{x+7}=8\end{cases}}}\)
\(TH1:5\sqrt{x+1}=\sqrt{x+7}\Leftrightarrow25\left(x-1\right)=x+7\Rightarrow x=\frac{4}{3}\)
\(TH2:5\sqrt{x-1}+\sqrt{x+7}=8\)
\(\Leftrightarrow5\sqrt{x-1}-5+\sqrt{x+7}-3=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{x-1}+1}+\frac{x-2}{\sqrt{x-7}+3}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x-7}+3}\right)=0\)
\(\Rightarrow x=2\)