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a) \(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2.2}=\sqrt{4}=2\)
b) \(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{8+2\sqrt{15}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)^2\left(\sqrt{15}-4\right)=2\left(4+\sqrt{15}\right)\left(\sqrt{15}-4\right)\)
\(=-2\)
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)
\(=5+3\sqrt{2}\)
d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=3\sqrt{2}+1\)
\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)
\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)
\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)
\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)
\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)
\(=\)\(3\sqrt{6}+9\)
Chúc bạn học tốt ~
\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=\)\(\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}\)
\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) )
\(=\)\(2-\sqrt{3}+\left|\sqrt{3}-1\right|\)
\(=\)\(2-\sqrt{3}+\sqrt{3}-1\) ( vì \(\sqrt{3}>\sqrt{1}=1\) )
\(=\)\(1\)
Chúc bạn học tốt ~
PS : mới lớp 8 sai thì thông cảm >.<
a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
Câu 2:
ĐKXĐ \(\hept{\begin{cases}x\ge0\\x-1\ne0\\x+2\sqrt{x}+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\\\left(\sqrt{x}+1\right)^2\ne0\end{cases}}\)
\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\frac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}=\frac{2x}{x-1}\)
\(4< 2\sqrt{15}\) nên căn thức k xác định
Bạn hình như chép nhầm đề rồi
\(4=\sqrt{16};2\sqrt{15}=\sqrt{60}\)
Sao trừ trong căn được