sqrt ( 12-3 )sqrt(7) - sqrt ( 12+3)sqrt(7 )
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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: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
27 tháng 7 2021
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
14 tháng 9 2021
\(\sqrt{7-2\sqrt{12}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
=> Chọn C
23 tháng 6 2023
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}\right)\)
=-2*căn 3/2=-căn 6
NQ
3
C
22 tháng 9 2019
\(\sqrt{12-3}.\sqrt{7}-\sqrt{12+3}.\sqrt{7}\)
\(=\sqrt{7}.\sqrt{12^2-3^2}\)
\(=\sqrt{7}.\sqrt{135}\)
\(=\sqrt{945}\)
22 tháng 9 2019
\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
Ta có :
\(\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)^2\)
= \(12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{12-3\sqrt{7}}.\sqrt{12+3\sqrt{7}}\)
= \(24-2.\sqrt{12^2-\left(3\sqrt{7}\right)^2}\)
= \(24-2.\sqrt{144-63}\)
= \(24-18=6\)
Mặt khác ta dễ thấy : \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}< 0\)
\(\Rightarrow\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=-\sqrt{6}\)
Chúc bạn học tốt !!!
NT
2
H9
HT.Phong (9A5)
CTVHS
19 tháng 9 2023
\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(A=\dfrac{\sqrt{2}\cdot\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)}{\sqrt{2}}\)
\(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{12+6\sqrt{7}}}{\sqrt{\text{2}}}\)
\(A=\dfrac{\sqrt{21-2\cdot\sqrt{21}\cdot\sqrt{3}+3}-\sqrt{21+2\cdot\sqrt{21}\cdot\sqrt{3}+3}}{\sqrt{2}}\)
\(A=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(A=\dfrac{\left|\sqrt{21}-\sqrt{3}\right|-\left|\sqrt{21}+\sqrt{3}\right|}{\sqrt{2}}\)
\(A=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{\text{2}}}\)
\(A=\dfrac{-\sqrt{6}}{\sqrt{2}}\)
\(A=-\sqrt{\dfrac{6}{2}}\)
\(A=-\sqrt{3}\)
ND
Nguyễn Đức Trí
VIP
19 tháng 9 2023
\(A=\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}\)
Giả sử \(\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}>0\)
\(\Leftrightarrow\sqrt[]{12-3\sqrt[]{7}}>\sqrt[]{12+3\sqrt[]{7}}\)
\(\Leftrightarrow12-3\sqrt[]{7}>12+3\sqrt[]{7}\)
\(\Leftrightarrow6\sqrt[]{7}< 0\left(sai\right)\)
Vậy \(\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}< 0\) hay \(A< 0\)
\(\Leftrightarrow A^2=12-3\sqrt[]{7}+12+3\sqrt[]{7}-2\sqrt[]{\left(12-3\sqrt[]{7}\right)\left(12+3\sqrt[]{7}\right)}\)
\(\Leftrightarrow A^2=24-2\sqrt[]{\left(144-63\right)}\)
\(\Leftrightarrow A^2=24-2\sqrt[]{81}\)
\(\Leftrightarrow A^2=24-18=6\)
\(\Leftrightarrow A=-\sqrt[]{6}\)
12 tháng 10 2021
c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)