Tính giá trị các biểu thức sau:
a) \(\sqrt{\frac{129}{16}+\sqrt{2}}\)
b) \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
c) \(\sqrt{2}.\sqrt{7-3\sqrt{5}}\)
d) \(\sqrt{\frac{59}{25}+\frac{6}{5}.\sqrt{2}}\)
e) \(\sqrt{2-\sqrt{3}.\left(\sqrt{6}+\sqrt{2}\right)}\)
K
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 6 2019
\(\sqrt{2-2.\frac{1}{2}\sqrt{2}+\frac{1}{4}}.\sqrt{8-2.2\sqrt{2}.\frac{1}{4}+\frac{1}{16}}=\sqrt{\left(\sqrt{2}-\frac{1}{2}\right)^2}\sqrt{\left(2\sqrt{2}-\frac{1}{4}\right)^2}\)
\(=\left(\sqrt{2}-\frac{1}{2}\right)\left(2\sqrt{2}-\frac{1}{4}\right)=\frac{33-10\sqrt{2}}{8}\)
\(\sqrt{2+2\sqrt{2}+1}.4\sqrt{\frac{288+2\sqrt{288}+1}{16}}=\sqrt{\left(\sqrt{2}+1\right)^2}.4\sqrt{\frac{\left(12\sqrt{2}+1\right)^2}{4^2}}\)
\(=\left(\sqrt{2}+1\right)\left(12\sqrt{2}+1\right)=25+13\sqrt{2}\)
\(\sqrt{28-10\sqrt{3}}=\sqrt{25-2.5\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
9 tháng 9 2021
a, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9}{4}-\frac{4\sqrt{2}}{4}}=\sqrt{\frac{9-4\sqrt{2}}{4}}\)
\(=\sqrt{\frac{\left(2\sqrt{2}\right)^2-4\sqrt{2}+1}{4}}=\sqrt{\frac{\left(2\sqrt{2}-1\right)^2}{4}}=\frac{2\sqrt{2}-1}{2}\)
b, \(\sqrt{\frac{129}{16}+\sqrt{2}}=\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}\right)^2+16\sqrt{2}+1}{16}}\)
\(=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)
23 tháng 7 2019
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
LD
2
11 tháng 6 2018
\(\sqrt{\frac{289+4\sqrt{72}}{16}}+\sqrt{\frac{129}{16}+\sqrt{2}}\)
\(=\sqrt{\frac{288+2\times12\sqrt{2}+1}{4^2}}+\sqrt{\frac{128+2\sqrt{12}+1}{4^2}}\)
\(=\sqrt{\frac{\left(\sqrt{288}+1\right)^2}{4^2}}+\sqrt{\frac{\left(\sqrt{128}+1\right)^2}{4^2}}\)
\(=\frac{\sqrt{288}+1}{4}+\frac{\sqrt{128}+1}{4}\)
\(=\frac{12\sqrt{2}+8\sqrt{2}+2}{4}\)
\(=\frac{1+10\sqrt{2}}{2}\)
26 tháng 10 2016
\(M=4\frac{1}{3}-\sqrt{16}+5\sqrt{\frac{4}{9}}-\frac{25}{\left(\sqrt{6}\right)^2}\)
\(=\frac{13}{3}-4+5\cdot\frac{2}{3}-\frac{25}{6}\)
\(=\frac{1}{3}+\frac{10}{3}-\frac{25}{6}\)
\(=\frac{11}{3}-\frac{25}{6}\)
\(=-\frac{1}{2}\)
15 tháng 8 2019
1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)
\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)
4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)
4A = \(8\sqrt{2}+1\)
⇒ A = \(\frac{\text{}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)
2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)
4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)
4B = \(12\sqrt{2}+1\)
⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)
3. \(\sqrt{2-\sqrt{3}}\). \(\left(\sqrt{6}+\sqrt{2}\right)\)
= \(\sqrt{2-\sqrt{3}}\). \(\sqrt{2}.\left(\sqrt{3}+1\right)\)
= \(\sqrt{4-2\sqrt{3}}\) . \(\left(\sqrt{3}+1\right)\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}\) . \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}-1\right)\). \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}\right)^2\) - 12
= 3 - 1
= 2
4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)
= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
= \(\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\) . \(\left(\sqrt{7}-\sqrt{3}\right)\)
= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)
= \(\sqrt{7}\) . (7 - 3)
= 4\(\sqrt{7}\)
5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{3+\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)
= \(\left(\sqrt{10}\right)^2-\left(\sqrt{2}\right)^2\)
= 10 - 2
= 8
6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)
= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)
= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30
= 2
7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)
= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)
= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)
= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)
= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)
TICK MÌNH NHA!
