A = \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
Tính A
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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
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: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
6 tháng 8 2019
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
12 tháng 10 2021
c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)
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: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
3 tháng 8 2017
a) A=12\(\sqrt{3}\)
B= \(\frac{8}{3}\)
c) C= 1
d)...
Chúc bạn học tốt nha ^^!
\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(A=\dfrac{\sqrt{2}\cdot\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)}{\sqrt{2}}\)
\(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{12+6\sqrt{7}}}{\sqrt{\text{2}}}\)
\(A=\dfrac{\sqrt{21-2\cdot\sqrt{21}\cdot\sqrt{3}+3}-\sqrt{21+2\cdot\sqrt{21}\cdot\sqrt{3}+3}}{\sqrt{2}}\)
\(A=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(A=\dfrac{\left|\sqrt{21}-\sqrt{3}\right|-\left|\sqrt{21}+\sqrt{3}\right|}{\sqrt{2}}\)
\(A=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{\text{2}}}\)
\(A=\dfrac{-\sqrt{6}}{\sqrt{2}}\)
\(A=-\sqrt{\dfrac{6}{2}}\)
\(A=-\sqrt{3}\)
\(A=\sqrt[]{12-3\sqrt[]{7}}-\sqrt[]{12+3\sqrt[]{7}}\)
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 \(A< 0\)
\(\Leftrightarrow A^2=12-3\sqrt[]{7}+12+3\sqrt[]{7}-2\sqrt[]{\left(12-3\sqrt[]{7}\right)\left(12+3\sqrt[]{7}\right)}\)
\(\Leftrightarrow A^2=24-2\sqrt[]{\left(144-63\right)}\)
\(\Leftrightarrow A^2=24-2\sqrt[]{81}\)
\(\Leftrightarrow A^2=24-18=6\)
\(\Leftrightarrow A=-\sqrt[]{6}\)