3-\(^{\dfrac{6}{7}^6}\)+\(\dfrac{1}{2}^2\):2
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Y
14 tháng 8 2023
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
BF
19 tháng 2 2023
`3/4 + 5/6 = 9/12 + 10/12 = 19/12`
`1/2 + 7/12 = 6/12 + 7/12 = 13/12`
`2/3 xx 3/4 = 2/4 = 1/2`
`7/4 : 2 = 7/4 xx 1/2 = 7/8`
VQ
19 tháng 2 2023
\(a,\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{18}{24}+\dfrac{20}{24}=\dfrac{38}{24}=\dfrac{19}{12}\)
\(b,\dfrac{1}{2}+\dfrac{7}{12}=\dfrac{6}{12}+\dfrac{7}{12}=\dfrac{13}{12}\)
\(c,\dfrac{2}{3}x\dfrac{3}{4}=\dfrac{2}{4}\)
\(d,\dfrac{7}{4}:2=\dfrac{7}{4}x\dfrac{1}{2}=\dfrac{7}{8}\)
C
17 tháng 4 2022
4/3 + 3/3 = 7/3
5/2 x 4 = 10
2 - 7/8 = 16/8 - 7/8 = 9/8
5 x 7/5 = 7
5/6 x 7/2= 35/12
17 tháng 4 2022
4/3 + 3/3 = 7/3
5/2 x 4 = 10
2 - 7/8 = 16/8 - 7/8 = 9/8
5 x 7/5 = 7
5/6 x 7/2= 35/12
9 tháng 1 2024
a: \(\dfrac{2}{5}+\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{-2}{3}+\dfrac{1}{2}\)
\(=\dfrac{2}{5}-\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{1}{2}\)
b: \(2,5-\left(-\dfrac{5}{6}\right)^0+\left(-\dfrac{1}{6}\right)^2\cdot\left(-3\right)\)
\(=\dfrac{5}{2}-1+\dfrac{1}{36}\cdot\left(-3\right)\)
\(=\dfrac{3}{2}-\dfrac{1}{12}=\dfrac{18}{12}-\dfrac{1}{12}=\dfrac{17}{12}\)
HT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 2 2023
a, \(\dfrac{x-1}{21}\) = \(\dfrac{3}{x+1}\)
( x-1)(x+1) = 21.3
x2 + x - x -1 = 63
x2 = 63 + 1
x2 = 64
x = + - 8
b, 2\(\dfrac{1}{2}\)x + x = 2\(\dfrac{1}{17}\)
x( \(\dfrac{5}{2}\) + 1) = \(\dfrac{35}{17}\)
x = \(\dfrac{35}{17}\) : ( \(\dfrac{5}{2}\)+1)
x = \(\dfrac{35}{17}\) x \(\dfrac{2}{7}\)
x = \(\dfrac{10}{17}\)
c, (x + \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\) ) : ( 2 + \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)): \(\dfrac{23}{12}\) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)) = \(\dfrac{7}{46}\) x \(\dfrac{23}{12}\)
x - \(\dfrac{5}{12}\) = \(\dfrac{7}{12}\)
x = \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)
x = 1
d, 2\(\dfrac{1}{3}\)x - 1\(\dfrac{3}{4}\)x + \(2\dfrac{2}{3}\) = 3\(\dfrac{3}{5}\)
x( \(\dfrac{7}{3}\) - \(\dfrac{7}{4}\)) + \(\dfrac{8}{3}\) = \(\dfrac{18}{5}\)
x\(\dfrac{7}{12}\) = \(\dfrac{18}{5}\) - \(\dfrac{8}{3}\)
x\(\dfrac{7}{12}\) = \(\dfrac{14}{15}\)
x = \(\dfrac{14}{15}\) : \(\dfrac{7}{12}\)
x = \(\dfrac{8}{5}\)
24 tháng 2 2022
a, \(=\dfrac{1+4}{5}+\dfrac{5+1+3}{9}=1+1=2\)
b, \(=\dfrac{1+4+2}{3}+\dfrac{1+2+5}{6}=\dfrac{6}{3}+\dfrac{8}{6}=2+\dfrac{4}{3}=\dfrac{6+4}{3}=\dfrac{10}{3}\)
12 tháng 10 2021
\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)
\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)
\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)
\(3-\dfrac{6^6}{7}+\dfrac{1^2}{2}\div2\)
\(=3-\dfrac{46656}{7}+\dfrac{1}{2}\times2\)
\(=4-\dfrac{46656}{7}\)
\(=\dfrac{28-46656}{7}\)
\(=\dfrac{-46628}{7}\)