Rút gọn A= 5√1÷5+1÷2√20+√5
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A) \(\dfrac{5}{10}-\dfrac{2}{15}\)
\(=\dfrac{1}{2}-\dfrac{2}{15}\)
\(=\dfrac{15}{30}-\dfrac{4}{30}\)
\(=\dfrac{11}{30}\)
B)\(\dfrac{5}{20}-\dfrac{1}{6}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}\)
\(=\dfrac{6}{24}-\dfrac{4}{24}\)
\(=\dfrac{2}{24}=\dfrac{1}{12}\)
C) \(\dfrac{6}{18}-\dfrac{6}{24}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}\)
\(=\dfrac{4}{12}-\dfrac{3}{12}\)
\(=\dfrac{1}{12}\)
D) \(\dfrac{5}{9}-\dfrac{3}{12}\)
\(=\dfrac{5}{9}-\dfrac{1}{4}\)
\(=\dfrac{20}{36}-\dfrac{9}{36}\)
\(=\dfrac{11}{36}\).
a: =1/2-2/15
=15/30-4/30
=11/30
b: =1/4-1/6
=3/12-2/12
=1/12
c: =1/3-1/4
=1/12
d: =20/36-9/36
=11/36
a) \(\dfrac{5}{6}-\dfrac{2}{14}\)
\(=\dfrac{5}{6}-\dfrac{1}{7}\)
\(=\dfrac{35}{42}-\dfrac{6}{42}\)
\(=\dfrac{29}{42}\)
b) \(\dfrac{5}{20}-\dfrac{1}{6}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}\)
\(=\dfrac{3}{12}-\dfrac{2}{12}\)
\(=\dfrac{1}{12}\)
c) \(\dfrac{5}{9}-\dfrac{3}{12}\)
\(=\dfrac{5}{9}-\dfrac{1}{4}\)
\(=\dfrac{20}{36}-\dfrac{9}{36}\)
\(=\dfrac{11}{36}\)
d) \(8-\dfrac{4}{6}\)
\(=\dfrac{48}{6}-\dfrac{4}{6}\)
\(=\dfrac{44}{6}=\dfrac{22}{3}\).
a, 5/10 - 2/15 = 1/2 - 2/15 = 15/30 - 4/30 = 11/30
b,5/20 - 1/6 = 1/4 - 1/6 = 3/12 - 2/12 = 1/12
c,6/18 - 6/24 = 1/3 - 1/4 = 4/12 - 3/12 = 1/12
d,5/9 - 3/12 = 5/9 - 1/4 = 20/36 - 9/36 = 11/36
chắc là đúng á , tại lớp 8 rồi ko nhớ lắm , mà vote mình nhá
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
Ta có A=1/5+1/20+1/44+...+1/170
=>A=1/10+1/40+1/88+...+1/340
A=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+......+\(\frac{1}{17.20}\)
A=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+.....+1/17-1/20)
A=1/3.(1/2-1/20)
A=\(\frac{3}{20}\)
`(5sqrt{1/5}+1/2sqrt{20}-5/4sqrt{4/5}+sqrt{5}):2/5
`=(sqrt5+1/2*2sqrt5-sqrt{5/4}+sqrt5):2/5`
`=(sqrt5+sqrt5+sqrt5-sqrt5/2):2/5`
`=(5/2*sqrt5):2/5`
`=25/4sqrt5`
`1/3sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1 1/3}`
`=1/3*4sqrt3+3*5sqrt3-3sqrt3-10sqrt{4/3}`
`=4/sqrt3+15sqrt3-3sqrt3-20/sqrt3`
`=12sqrt3-16/sqrt3`
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
a,
B = 1 + 5 + 5^2 + 5^3 + ... + 5^100
5B = 5 + 5^2 + 5^3 + ... + 5^101
5B - B = [5 + 5^2 + 5^3 + ... + 5^101] - [1 + 5 + 5^2 + 5^3 + ... + 5^100]
4B = 5 + 5^2 + 5^3 + ... + 5^101 - 1 - 5 - 5^2 - 5^3 - ... - 5^100
4B = 5^101 - 1
B = \(\frac{5^{101}-1}{4}\)
b,
A = 1 - 3 + 3^2 - 3^3 + ... + 3^20 - 3^21
3A = 3 - 3^2 + 3^3 - 3^4 + ... + 3^21 - 3^22
3A - A = [3 - 3^2 + 3^3 - 3^4 + ... + 3^21 - 3^22] - [1 - 3 + 3^2 - 3^3 + ... + 3^20 - 3^21]
2A = 3 - 3^2 + 3^3 - 3^4 + ... + 3^21 - 3^22 - 1 + 3 - 3^2 + 3^3 + ... - 3^20 + 3^21
2A = 2[3 + 3^3 + 3^5 + ... + 3^21] - 2[3^2 + 3^4 + ... + 3^20] - 1
Đặt C = 3 + 3^3 + 3^5 + ... + 3^21
=> 3^2C = 3^3 + 3^5 + 3^7 + ... + 3^23
=> 9C - C = [3^3 + 3^5 + 3^7 + ... + 3^23] - [3 + 3^3 + 3^5 + ... + 3^21]
=> 8C = 3^3 + 3^5 + 3^7 + ... + 3^23 - 3 - 3^3 - 3^5 - ... - 3^21
=> 8C = 3^23 - 3
=> C = 3^23 - 3 / 8
=> 2[3 + 3^3 + 3^5 + ... + 3^21] = 3^23 - 3 / 8 * 2 = 3^23 - 3 / 4
Đặt D = 3^2 + 3^4 + ... + 3^20
=> 3^2D = 3^4 + 3^6 + ... + 3^22
=> 9D - D = [3^4 + 3^6 + ... + 3^22] - [3^2 + 3^4 + ... + 3^20]
=> 8D = 3^4 + 3^6 + ... + 3^22 - 3^2 - 3^4 - ... - 3^20
=> 8D = 3^22 - 9
=> D = 3^22 - 9 / 8
=> 2[3^2 + 3^4 + ... + 3^20] = 3^22 - 9 / 8 * 2 = 3^22 - 9 / 4
=> A = 3^23 - 3 / 4 - 3^22 - 9 / 4 - 1
\(\Rightarrow A=\frac{3^{23}-3-3^{22}-9}{4}-1=\frac{3^{22}\left[3-1\right]-12}{4}=\frac{3^{22}\cdot2-12}{4}\)
\(=\frac{6\left[3^{21}-2\right]}{4}=\frac{3\left[3^{21}-2\right]}{2}=5230176601\)
Mình chỉ biết làm thế thôi, sai thì mong mn sửa lại giúp nhé
\(A=\frac{5\sqrt{1}}{5}+\frac{1}{2}\sqrt{20}+\sqrt{5}=\sqrt{1}+\frac{1}{2}\cdot2\sqrt{5}+\sqrt{5}=\sqrt{1}+2\sqrt{5}=1+2\sqrt{5}\)