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AH
Akai Haruma
Giáo viên
18 tháng 7 2023
Lời giải:
\(A=\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
$A\sqrt{2}=\sqrt{16-6\sqrt{7}}+\sqrt{8-2\sqrt{7}}$
$=\sqrt{(3-\sqrt{7})^2}+\sqrt{(\sqrt{7}-1)^2}$
$=|3-\sqrt{7}|+|\sqrt{7}-1|$
$=3-\sqrt{7}+\sqrt{7}-1=2$
11 tháng 6 2018
\(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}\right)^2-2^2}-\frac{5\left(4-\sqrt{7}\right)}{4^2-\left(\sqrt{7}\right)^2}\)
\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{8}\)
\(=\frac{12\left(\sqrt{7}-5\right)}{24}-\frac{6\left(6+2\sqrt{7}\right)}{24}+\frac{8\left(6\sqrt{7}+12\right)}{24}-\frac{3\left(20-5\sqrt{7}\right)}{24}\)
\(=\frac{12\sqrt{7}-60-36-12\sqrt{7}+48\sqrt{7}+96-60+15\sqrt{7}}{24}\)
\(=\frac{-60+63\sqrt{7}}{24}\)
5 tháng 9 2023
a) \(\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)\)
\(=\sqrt{2\cdot\left(4+\sqrt{7}\right)}+\sqrt{2\cdot\left(4-\sqrt{7}\right)}\)
\(=\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=\left|\sqrt{7}+1\right|+\left|\sqrt{7}-1\right|\)
\(=\sqrt{7}+1+\sqrt{7}-1\)
\(=2\sqrt{7}\)
b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2\cdot\left(2-\sqrt{3}\right)}-\sqrt{2\cdot\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{ }\)
\(=-\dfrac{2}{\sqrt{2}}\)
\(=-\sqrt{2}\)
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
6 tháng 8 2019
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
NA
1
23 tháng 10 2019
Ta có:
\(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(\sqrt{2}A=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(\sqrt{2}A=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\sqrt{7}-1-\sqrt{7}-1=-2\)
\(\Rightarrow A=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
`@` `\text {Ans}`
`\downarrow`
`\sqrt {7} + 4 \sqrt {3} - \sqrt {7} - 4 \sqrt {3}`
`= ( \sqrt {7} - \sqrt {7}) + ( 4 \sqrt {3} - 4 \sqrt {3})`
`= 0 + 0`
`= 0`