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a) \(A=\frac{3}{\sqrt{7}-2}+\frac{7}{\sqrt{7}-\sqrt{28}}=\frac{3}{\sqrt{7}-2}-\sqrt{7}\)
\(=\frac{3-7+2\sqrt{7}}{\sqrt{7}-2}=\frac{-4+2\sqrt{7}}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
đk: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(B=\left(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(B=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(B=\frac{2}{\sqrt{x}+2}\)
b) \(6B-A>0\Leftrightarrow\frac{12}{\sqrt{x}+2}-2>0\)
\(\Leftrightarrow\frac{8-2\sqrt{x}}{\sqrt{x}+2}>0\Rightarrow8-2\sqrt{x}>0\left(because:\sqrt{x}+2>0\right)\)
\(\Rightarrow\sqrt{x}< 4\Rightarrow x< 16\)
Vậy \(\hept{\begin{cases}0< x< 16\\x\ne4\end{cases}}\)
Đặt \(A=\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\)
\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\left(\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\right)\)
\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\cdot A\)
\(\Rightarrow A^3=50-3A\sqrt[3]{343}=50-21A\)
\(\Rightarrow A^3+21A-50=0\Leftrightarrow A^3-4A+25A-50=0\)
\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+25\right)=0\)
\(\Leftrightarrow A=2\left(A^2+2A+25>0,\forall A\right)\)
\(\Rightarrow\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}=2\)
Tick nha bạn 😘
a, Ta có :\(2.x^2-2.x+0,5=0\)
\(\Leftrightarrow4.x^2-4.x+1=0\) (Nhân mỗi vế với 2)
\(\Leftrightarrow\) \(\left(2.x-1\right)^2=0\)
\(\Leftrightarrow2.x-1=0\)
\(\Leftrightarrow2.x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
STUDY WELL!