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D
1 tháng 12 2017
1) \(\sqrt{36+12\sqrt{5}}=\sqrt{\left(\sqrt{30}+\sqrt{6}\right)^2}=\sqrt{30}+\sqrt{6}\)
2)\(\sqrt{21-6\sqrt{6}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=\sqrt{18}-\sqrt{3}\)
3)\(\sqrt{6-2\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{9}-1\right)^2}\)
\(=\sqrt{5}-1-\left(\sqrt{9}-1\right)\)
\(=\sqrt{5}-\sqrt{9}\)
4)\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1-\left(\sqrt{2-1}\right)=2\)
5) \(\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)=2\sqrt{3}\)
6)\(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-\left(3-\sqrt{2}\right)=2\sqrt{2}-1\)
7)\(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}=\sqrt{\left(\sqrt{20}-1\right)^2}+\sqrt{\left(\sqrt{20}+1\right)^2}\)
\(=\sqrt{20}-1+\sqrt{20+1}=2\sqrt{20}\)
16 tháng 5 2019
câu b:
(\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\))^2
\(=\left(5+2\sqrt{6}\right)-\left(5-2\sqrt{6}\right)\)\(-2\sqrt{5+2\sqrt{6}}\sqrt{5-2\sqrt{6}}\)
\(=4\sqrt{6}-2\sqrt{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}\)
\(=4\sqrt{6}-2\sqrt{5^2-\left(2\sqrt{6}\right)^2}\)
\(=4\sqrt{6}-2\sqrt{25-24}=4\sqrt{6}-2\)
mấy câu khác tương tự
18 tháng 7 2020
a) Ta có: \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)
b) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)
c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|3+\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}+3+\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3+2\sqrt{3}+2\sqrt{2}\)
h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)
\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)
\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)
\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)
\(=7-\sqrt{45}-7-\sqrt{45}\)
\(=-2\sqrt{45}=-6\sqrt{5}\)
i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\cdot\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A=\sqrt{5}+1\)
AH
Akai Haruma
Giáo viên
5 tháng 8 2020
2.1
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5.1}+1}-\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=2\)
2.2
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{3+2\sqrt{3.5}+5}+\sqrt{3-2\sqrt{3.5}+5}-2\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-2\sqrt{(\sqrt{5}-1)^2}\)
\(=|\sqrt{3}+\sqrt{5}|+|\sqrt{3}-\sqrt{5}|-2|\sqrt{5}-1|=2\)
$\Rightarrow B=\sqrt{2}$
AH
Akai Haruma
Giáo viên
5 tháng 8 2020
Bài 1:
1. ĐKXĐ: \(\left\{\begin{matrix} 2x-1\geq 0\\ x-3\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x\geq 3\\ x< 5\end{matrix}\right.\Leftrightarrow 3\leq x< 5\)
2.
ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 2-x\geq 0\\ x+1>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 2\\ x>-1\end{matrix}\right.\Leftrightarrow 1\leq x\leq 2\)
22 tháng 7 2017
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
KT
4 tháng 7 2018
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
\(\sqrt{5^{2^{ }}-4^2}\) = \(\sqrt{25-16}\) = \(\sqrt{9}\) = 3
\(\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3\)