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c: \(=\sqrt{\dfrac{4}{16-6\sqrt{7}}}+\sqrt{7}\)
\(=\dfrac{2}{3-\sqrt{7}}+\sqrt{7}\)
\(=3+2\sqrt{7}\)
d: \(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
4) Ta có pt \(\Leftrightarrow\dfrac{7x+1+x^2-8x-1}{\sqrt[3]{\left(7x+1\right)^2}-\sqrt[3]{\left(7x+1\right)\left(x^2-8x-1\right)}+\sqrt[3]{\left(x^2-8x+1\right)^2}}+\dfrac{x^2-x+8-8}{\sqrt[3]{\left(x^2-x+8\right)^2}+2\sqrt[3]{x^2-x+8}+4}=0\)
\(\Leftrightarrow\dfrac{x^2-x}{...}+\dfrac{x^2-x}{...}=0\Leftrightarrow\left(x^2-x\right)\left(...\right)=0\)
Mà ...>0 => \(x^2-x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
2) Ta có pt \(\Leftrightarrow\sqrt{x\left(x+1\right)}-\sqrt{x-1}=\sqrt{x}\Leftrightarrow x\left(x+1\right)=\left(\sqrt{x}+\sqrt{x-1}\right)^2\)
\(\Leftrightarrow x^2+x=2x-1+2\sqrt{x\left(x-1\right)}\Leftrightarrow x^2-x-1=2\left(\sqrt{x^2-x}-1\right)\)
\(\Leftrightarrow x^2-x-1=2.\dfrac{x^2-x-1}{\sqrt{x^2-x}+1}\Leftrightarrow\left(x^2-x-1\right)\left(1-\dfrac{2}{\sqrt{x^2-x}+1}\right)=0\)...đến đấy chắc tự làm tiếp được
2. ĐK: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x}\ge0\\b=\sqrt{x^2+4}\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=2a^2\\x^2+4=b^2\\3\sqrt{x^3+4x}=3ab\end{matrix}\right.\)
pt trên được viết lại thành
\(2a^2+b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=\dfrac{1}{2}b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{x^2+4}\\\sqrt{x}=\dfrac{1}{2}\sqrt{x^2+4}\end{matrix}\right.\)
Đến đây dễ rồi nhé ^^
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
\(< =>\sqrt[3]{x+5}=-2\)
<=> \(\left(\sqrt[3]{x+5}\right)^3=-8\)
<=> \(x+5=-8\)
<=> x=-13
a) \(\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}=4\) (1)
\(\Leftrightarrow\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-4=0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}+2+\sqrt{x}-4\left(2+\sqrt{x}\right)\cdot\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=0\)
\(\Leftrightarrow2-\sqrt{x}+2+\sqrt{x}-4\left(2+\sqrt{x}\right)\cdot\left(2-\sqrt{x}\right)=0\)
\(\Leftrightarrow2+2-4\left(4-x\right)=0\)
\(\Leftrightarrow2+2-16+4x=0\)
\(\Leftrightarrow-12+4x=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{3\right\}\)
b) \(\dfrac{8-\sqrt{x}}{\sqrt{x}-7}+\dfrac{1}{7-\sqrt{x}}=8\) (2)
\(\Leftrightarrow\dfrac{8-\sqrt{x}}{\sqrt{x}-7}+\dfrac{1}{7-\sqrt{x}}-8=0\)
\(\Leftrightarrow\dfrac{8-\sqrt{x}-1-8\left(\sqrt{x}-7\right)}{\sqrt{x}-7}=0\)
\(\Leftrightarrow8-\sqrt{x}-1-8\left(\sqrt{x}-7\right)=0\)
\(\Leftrightarrow8-\sqrt{x}-1-8\sqrt{x}+56=0\)
\(\Leftrightarrow63-9\sqrt{x}=0\)
\(\Leftrightarrow-9\sqrt{x}=-63\)
\(\Leftrightarrow\sqrt{x}=7\)
\(\Leftrightarrow x=49\)
sau khi thử lại ta nhận thấy: \(\dfrac{8-\sqrt{49}}{\sqrt{49}-8}+\dfrac{1}{7-\sqrt{49}}=8\)\(\Leftrightarrow\dfrac{1}{0}+\dfrac{1}{7-\sqrt{49}}=8\)
\(\Rightarrow x\ne48\)
\(\Rightarrow x\in\varnothing\)