Rút gọn: \(\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}\)
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9 tháng 10 2021
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
17 tháng 1 2022
, \(A=\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
\(=\frac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{3}-\frac{5\left(4-\sqrt{7}\right)}{9}\)
\(=\frac{-16+4\sqrt{7}}{4}+\frac{18\sqrt{7}+36-20+5\sqrt{7}}{9}=-4+\sqrt{7}+\frac{23\sqrt{7}+16}{9}\)
b,\(B=\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-2\right)}{2}+\frac{5\sqrt{6}}{6}\)
\(=\frac{12\sqrt{6}+5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
AH
Akai Haruma
Giáo viên
11 tháng 8 2021
Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
AH
Akai Haruma
Giáo viên
29 tháng 12 2023
Bài 1:
a/
$\sqrt{(\sqrt{7}-4)^2}+\sqrt{8-2\sqrt{7}}$
$=|\sqrt{7}-4|+\sqrt{7+1-2\sqrt{7}}=|\sqrt{7}-4|+\sqrt{(\sqrt{7}-1)^2}$
$=4-\sqrt{7}+|\sqrt{7}-1|=4-\sqrt{7}+\sqrt{7}-1=3$
b/
\(\sqrt{(\sqrt{5}-2)^2}+\sqrt{6+2\sqrt{5}}\\ =|\sqrt{5}-2|+\sqrt{5+1+2\sqrt{5}}\\ =\sqrt{5}-2+\sqrt{(\sqrt{5}+1)^2}\\ =\sqrt{5}-2+|\sqrt{5}+1|=\sqrt{5}-2+\sqrt{5}+1=2\sqrt{5}-1\)
AH
Akai Haruma
Giáo viên
29 tháng 12 2023
Bài 2:
a. $=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}$
b. $=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}$
$=\frac{\sqrt{2}+3\sqrt{2}+5\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}$
c.
$=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}$
$=-\sqrt{5}+15\sqrt{2}$
d.
$=0,1.10\sqrt{2}+2.\frac{\sqrt{2}}{5}+0,4.5\sqrt{2}$
$=\sqrt{2}+0,4\sqrt{2}+2\sqrt{2}$
$=\sqrt{2}(1+0,4+2)=3,4\sqrt{2}$
10 tháng 5 2017
\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(A=\sqrt{9}-\sqrt{1}=3-1=2\)
24 tháng 8 2023
a: Sửa đề: \(\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{3}-2}=\dfrac{2-\sqrt{3}}{\sqrt{3}-2}\)
=-1
b: Sửa đề: \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
=1
HP
31 tháng 8 2021
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
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: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
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/
\(=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)
\(\dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}=\dfrac{12\sqrt{6}}{\sqrt{\left(\sqrt{6}+1\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}}=\dfrac{12\sqrt{6}}{\sqrt{6}+1-\sqrt{6}+1}=\dfrac{12\sqrt{6}}{2}=6\sqrt{6}\)
\(=\dfrac{12\sqrt{6}}{\sqrt{6+2.\sqrt{6}.1+1}-\sqrt{6-2.\sqrt{6}.1+1}}\)\(=\dfrac{12\sqrt{6}}{\sqrt{6}+1-\sqrt{6}+1}=\dfrac{12\sqrt{6}}{2}=6\sqrt{6}\)