\(\sqrt{\dfrac{\left(2-\sqrt{5}\right)^2}{8}}\)
\(\dfrac{7}{3\sqrt[]{14}}\)
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19 tháng 8 2021
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
19 tháng 8 2021
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
18 tháng 7 2021
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
T
10 tháng 9 2023
a, \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\left|2-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot3+3^2}\)
\(=\sqrt{5}-2+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(=\sqrt{5}-2+\left|\sqrt{5}-3\right|\)
\(=\sqrt{5}-2+3-\sqrt{5}\)
\(=1\)
b, (ĐKXĐ: x ≥ 0; x ≠ 1)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{x-\sqrt{x}+3\sqrt{x}-3}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-5}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
#\(Toru\)
10 tháng 9 2023
a: \(=\sqrt{5}-2+3-\sqrt{5}=3-2=1\)
b:
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{x-5+\sqrt{x}-1+2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+\sqrt{x}-6+2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
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.
$=0,1.10\sqrt{2}+2.\frac{\sqrt{2}}{5}+0,4.5\sqrt{2}$
$=\sqrt{2}+0,4\sqrt{2}+2\sqrt{2}$
$=\sqrt{2}(1+0,4+2)=3,4\sqrt{2}$
24 tháng 11 2021
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
AH
Akai Haruma
Giáo viên
17 tháng 9 2021
Lời giải:
a.
\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)
b.
\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)
\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)
\(=10\)
AH
Akai Haruma
Giáo viên
17 tháng 9 2021
c.
\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)
\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)
d.
\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)
\(=|2+\sqrt{3}|-|5-\sqrt{3}|=2+\sqrt{3}-(5-\sqrt{3})=-3+2\sqrt{3}\)
6 tháng 7 2021
a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)
\(=10\sqrt{5}\)
b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)
\(=6-2\sqrt{3}+4+3\sqrt{3}\)
\(=10+\sqrt{3}\)
c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=7-5=2
d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=-3+2\sqrt{3}\)
6 tháng 7 2021
a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)
\(=10\sqrt{5}\)
b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)
\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)
\(=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)
\(=10\)
c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=7-5=2\)
d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|-\left(5-\sqrt{3}\right)^2\)
\(=\left|2+\sqrt{3}\right|-\left|5-\sqrt{3}\right|\)
\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=2\sqrt{3}-3\)
`\sqrt(((2-\sqrt5)^2)/8)`
`= (\sqrt((2-\sqrt5)^2))/(\sqrt8)`
`= (|2-\sqrt5|)/(2\sqrt2)`
`=(\sqrt5-2)/(2\sqrt2)`
`=(\sqrt10-2\sqrt2)/4`
.
`7/(3\sqrt14) = (\sqrt7 .\sqrt7)/(3.\sqrt7 .\sqrt2)`
`=(\sqrt7)/(3\sqrt2)`
`=(\sqrt14)/(3.2)`
`=(\sqrt14)/6`
\(\sqrt{\dfrac{\left(2−\sqrt{5}\right)^2}{8}}\)= \(\dfrac{\sqrt{5}-2}{2\sqrt{2}}\)
\(\dfrac{7}{3\sqrt{14}}\) = \(\dfrac{\sqrt{7}}{3\sqrt{2}}\)