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T
0
TT
2
8 tháng 8 2022
\(=\dfrac{\sqrt{13+2\sqrt{12}}-\sqrt{13-2\sqrt{12}}+2\sqrt{12}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{12}+1-\sqrt{12}+1+2\sqrt{12}}{\sqrt{2}}\)
\(=\dfrac{2\sqrt{12}+2}{\sqrt{2}}=2\sqrt{6}+\sqrt{2}\)
AH
Akai Haruma
Giáo viên
24 tháng 6 2018
Lời giải:
1)
Để biểu thức có nghĩa thì:
\(2x^2-5x+3\geq 0\)
\(\Leftrightarrow 2x(x-1)-3(x-1)\geq 0\)
\(\Leftrightarrow (2x-3)(x-1)\geq 0\)
\(\Leftrightarrow \left[\begin{matrix} x\geq \frac{3}{2}\\ x\leq 1\end{matrix}\right.\)
2)
\(\sqrt{6.5+\sqrt{12}}+\sqrt{6.5-\sqrt{12}}+2\sqrt{6}\)
\(=\sqrt{(\sqrt{6})^2+(\frac{1}{\sqrt{2}})^2+2\sqrt{6}.\frac{1}{\sqrt{2}}}+\sqrt{(\sqrt{6})^2+(\frac{1}{\sqrt{2}})^2-2\sqrt{6}.\frac{1}{\sqrt{2}}}+2\sqrt{6}\)
\(=\sqrt{(\sqrt{6}+\frac{1}{\sqrt{2}})^2}+\sqrt{(\sqrt{6}-\frac{1}{\sqrt{2}})^2}+2\sqrt{6}\)
\(=\sqrt{6}+\frac{1}{\sqrt{2}}+\sqrt{6}-\frac{1}{\sqrt{2}}+2\sqrt{6}=4\sqrt{6}\)
T
30 tháng 7 2018
c)
\(\sqrt{2}C=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)
\(=\sqrt{5}+1-\left(\sqrt{5}-1\right)-2=0\Rightarrow C=0\)
b)
\(B=3\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)\)
\(\Rightarrow\sqrt{2}B=3\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\)
\(=3\left(\sqrt{5}+1+\sqrt{5}-1\right)-\sqrt{5}\left(\sqrt{5}+1-\sqrt{5}+1\right)\)
\(\sqrt{2}B=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\Rightarrow B=2\sqrt{10}\)
24 tháng 6 2021
C)√3+√5−√3−√5−√2b) (3−√5)√3+√5+(3+√5)√3−√5d) √4−√7−√4+√7+√7e) √6,5+√12+√6,5−√12+2√6mình cần giải gấp ạ
HS
11 tháng 9 2018
\(A=\dfrac{5.\left(38^2-17^2\right)}{8\left(47^2-19^2\right)}\\ =\dfrac{5\left(38-17\right)\left(38+17\right)}{8\left(47-19\right)\left(47+19\right)}\\ =\dfrac{5.21.55}{8.28.66}\\ =\dfrac{5.1155}{8.1848}\\ =\dfrac{5.5}{8.8}\\ =\dfrac{25}{64}\)
\(B=\sqrt{\dfrac{0,2\times1,21\times0,3}{7,5\times3,2\times0,64}}\\ =\sqrt{0,0625\times1,890625\times0,04}\\ =\sqrt{\dfrac{121}{25600}}\\ =\dfrac{11}{160}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 3 2020
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)
\(B=\frac{1}{\sqrt{2}}\left(\left(3-\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\right)\)
\(=\frac{1}{\sqrt{2}}\left(\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)\right)\)
\(=\frac{1}{\sqrt{2}}\left(\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\right)\)
\(=\frac{1}{\sqrt{2}}\left(2\sqrt{5}-2+2\sqrt{5}+2\right)=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
\(C=\frac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{5}+1-\sqrt{5}+1-2\right)=0\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 3 2020
\(D=\frac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{14}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1+\sqrt{14}\right)\)
\(=\frac{1}{\sqrt{2}}\left(-2+\sqrt{14}\right)=\sqrt{7}-\sqrt{2}\)
\(E=\frac{1}{\sqrt{2}}\left(\sqrt{13+2\sqrt{12}}+\sqrt{13-2\sqrt{12}}\right)+2\sqrt{6}\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{12}+1\right)^2}+\sqrt{\left(\sqrt{12}-1\right)^2}\right)+2\sqrt{6}\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{12}+1+\sqrt{12}-1\right)+2\sqrt{6}\)
\(=\sqrt{24}+2\sqrt{6}=4\sqrt{6}\)
17 tháng 8 2021
a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)
c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)
d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)
e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)
f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)
AH
Akai Haruma
Giáo viên
4 tháng 9 2023
Lời giải:
a. \(=|\sqrt{7}-5|+|2-\sqrt{7}|=5-\sqrt{7}+(\sqrt{7}-2)=3\)
b. \(=\sqrt{(3+\sqrt{2})^2}-\sqrt{(3-\sqrt{2})^2}=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=(3+\sqrt{2})-(3-\sqrt{2})=2\sqrt{2}\)
c.
\(=\sqrt{(3+2\sqrt{2})^2}+\sqrt{(3-2\sqrt{2})^2}=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
$=(3+2\sqrt{2})+(3-2\sqrt{2})=6$
d.
$=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$
$=|\sqrt{5}+1|-|\sqrt{5}-1|=\sqrt{5}+1-(\sqrt{5}-1)=2$
Đặt \(A=\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\)
<=> \(A^2=\left(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\right)^2\)
<=> \(A^2=6,5+\sqrt{12}+2\sqrt{\left(6,5+\sqrt{12}\right)\left(6,5-\sqrt{12}\right)}+6,5-\sqrt{12}\)
<=> \(A^2=13+2\sqrt{42,25-12}\)
<=> \(A^2=13+2\sqrt{\frac{121}{4}}\)
<=> \(A^2=13+2\cdot\frac{11}{2}=13+11=24\)
=> \(A=2\sqrt{6}\)
Vậy \(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}+2\sqrt{6}=4\sqrt{6}\)