Rút gọn căn thức
\(\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\)
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
K
Khách
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Những câu hỏi liên quan
12 tháng 10 2023
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
28 tháng 5 2022
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
22 tháng 7 2016
a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
AH
Akai Haruma
Giáo viên
11 tháng 8 2021
Lời giải:
1/
\(=\frac{3.\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)
2/
\(=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)
3/
\(=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)
6 tháng 9 2023
\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{7\left(2+\sqrt{3}\right)}{4-3}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)
\(=4\left(\sqrt{5}+\sqrt{3}\right)-14-7\sqrt{3}+4\sqrt{2}+4+\sqrt{3}-1\)
\(=4\sqrt{5}+4\sqrt{3}-6\sqrt{3}+4\sqrt{2}-11\)
\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)
MH
6 tháng 9 2023
\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\dfrac{7\left(\sqrt{3}+2\right)}{3-4}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)
\(=4\sqrt{5}+4\sqrt{3}-7\sqrt{3}-14+4\sqrt{2}+4+\sqrt{3}-1\)
\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)
KT
2 tháng 7 2018
a) \(A=\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\)\(\sqrt{2}A=\sqrt{8-2\sqrt{15}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\left(\sqrt{3}+1\right)=\sqrt{5}-1\)
\(\Rightarrow\)\(A=\frac{\sqrt{5}-1}{\sqrt{2}}\)
b) tương tự câu a
c) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}-\sqrt{6-2\sqrt{5+\sqrt{\left(\sqrt{12}-1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}-\sqrt{6-2\sqrt{5+\left(\sqrt{12}-1\right)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}-\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}-\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}-\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
\(A=\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\)
\(\Rightarrow A\sqrt{2}=\sqrt{8+2\sqrt{15}}-\sqrt{14-2\sqrt{45}}\)
\(A\sqrt{2}=\sqrt{\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2.\sqrt{3}.\sqrt{5}}-\sqrt{\left(\sqrt{5}\right)^2+\left(\sqrt{9}\right)^2-2.\sqrt{5}.\sqrt{9}}\)
\(A\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{9}-\sqrt{5}\right)^2}\)
\(A\sqrt{2}=\sqrt{3}+\sqrt{5}-\sqrt{9}+\sqrt{5}=2\sqrt{5}+\sqrt{3}-\sqrt{9}\Rightarrow A=\frac{2\sqrt{5}+\sqrt{3}-\sqrt{9}}{\sqrt{2}}\)\(B=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\Rightarrow B\sqrt{2}=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(B\sqrt{2}=\sqrt{1^2+\left(\sqrt{3}\right)^2+2.1.\sqrt{3}}+\sqrt{\left(\sqrt{3}\right)^2+1^2-2.1.\sqrt{3}}\)\(B\sqrt{2}=\sqrt{\left(1+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=1+\sqrt{3}+\sqrt{3}-1=2\sqrt{3}\)