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A=1+5+5^2+5^3+...+5^2015
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CC
29 tháng 2 2020
\(A=1+5+5^2+5^3+...+5^{2015}\)
\(5A=5.\left(1+5+5^2+5^3+...+5^{2015}\right)\)
\(=5+5^2+5^3+5^4+...+5^{2016}\)
\(5A-A=\left(5+5^2+5^3+5^4+...+5^{2016}\right)-\left(1+5+5^2+5^3+...+5^{2015}\right)\)
\(4A=5^{2016}-1\)
\(A=\frac{5^{2016}-1}{4}\)
HT
1
LM
Lê Minh Vũ
VIP
9 tháng 8 2023
a) \(18+25\times4-4^3\)
\(=18+100-64\)
\(=118-64\)
\(=54\)
b) \(275-\left(49+125\div5^3\right)\)
\(=275-\left(49+125\div125\right)\)
\(=275-\left(49+1\right)\)
\(=275-49-1\)
\(=226-1\)
\(=225\)
c) \(2015+\left(8\times15-\left(18-8\right)^2\right)\)
\(=2015+\left(8\times15-\left(10\right)^2\right)\)
\(=2015+\left(8\times15-100\right)\)
\(=2015+\left(120-100\right)\)
\(=2015+20\)
\(=2035\)
C
10 tháng 12 2023
1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
26 tháng 12 2016
= 2015 + 5 x [300-100]
= 2015 + 5 x 200
= 2015 + 1000
= 3015
18 tháng 1 2022
\(a,\dfrac{5}{16}+\dfrac{-5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\\ b,\dfrac{-2}{7}+\dfrac{3}{-5}=\dfrac{-10}{35}-\dfrac{21}{35}=\dfrac{-10-21}{35}=\dfrac{-31}{35}\\ c,\dfrac{-2}{-5}+\dfrac{-5}{-6}=\dfrac{2}{5}+\dfrac{5}{6}=\dfrac{12}{30}+\dfrac{25}{30}=\dfrac{12+25}{30}=\dfrac{37}{30}\\ d,\dfrac{-5}{-3}+\dfrac{3}{-7}=\dfrac{5}{3}-\dfrac{3}{7}=\dfrac{35}{21}-\dfrac{9}{21}=\dfrac{35-9}{21}=\dfrac{26}{21}\)
NN
14 tháng 6 2020
a) \(x+5=2015-\left(12-7\right)\)
\(\Leftrightarrow x+5=2015-5\)\(\Leftrightarrow x+5=2010\)
\(\Leftrightarrow x=2005\)
Vậy \(x=2005\)
b) \(-7\left|x+3\right|=-49\)\(\Leftrightarrow\left|x+3\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=-7\\x+3=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-10\\x=4\end{cases}}\)
Vậy \(x=-10\)hoặc \(x=4\)
c) \(\left(105-x\right):2^5=2015^0+1\)
\(\Leftrightarrow\left(105-x\right):32=1+1\)\(\Leftrightarrow\left(105-x\right):32=2\)
\(\Leftrightarrow105-x=64\)\(\Leftrightarrow x=41\)
Vậy \(x=41\)
14 tháng 6 2020
a. x+5 = 2015-(12-7)
x+5= 2015-5
x+5 = 2010
x=2005
b, -7/ x+3/ =-49
/x+3/ = 7
x+3 = 7 suy ra x=4
hoặc x+3= -7 suy ra x=-10
c. (105-x) : 2^5= 2015^0 + 1
(105-x) : 32= 1+1
(105-x) : 32= 2
105-x= 64
x= 41
Cái này kiến thức cơ bản thôi ạ
XO
31 tháng 1 2020
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1
= [(2015 - 1) : 2 + 1].(2015 + 1) : 2
= 1008.2016 : 2 = 1016064
Lại có : 2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng
= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)
= 2 + 2 + ... + 2 + 2 (504 số hạng 2)
= 2 x 504 = 1008
Khi đó A = \(\frac{1016064}{1008}=1008\)
b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)
Khi đó : B/100 = 1/2
=> B = 50
Vậy B = 50
H9
HT.Phong (9A5)
CTVHS
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
TA CÓ:
\(A=1+5+5^2+....+5^{2015}\)
\(5A=5+5^2+5^3+...+5^{2016}\)
\(5A-A=\left(5+5^2+5^3+...+5^{2016}\right)-\left(1+5+5^2+...+5^{2015}\right)\)
\(4A=5^{2016}-1\)
\(A=\frac{5^{2016}-1}{4}\)
Vậy A=...