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Xét : \(\frac{a}{5}=\frac{12}{144}\Leftrightarrow a=\frac{5}{12}\)
Xét : \(\frac{b}{3}=\frac{12}{144}\Leftrightarrow b=\frac{1}{4}\)
Xét ; \(\frac{c}{8}=\frac{12}{144}\Leftrightarrow c=\frac{2}{3}\)
Ta có
\(\hept{\begin{cases}\frac{a}{5}=\frac{12}{144}\\\frac{b}{3}=\frac{12}{144}\\\frac{c}{8}=\frac{12}{144}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5.12:144=\frac{5}{12}\\b=3.12:144=\frac{1}{4}\\c=8.12:144=\frac{2}{3}\end{cases}}\)
vậy a=5/12 và b=1/4 và c=2/3
\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)
a) ta có: A = 3^0 + 3^1 + 3^2 + ...+ 3^100
=> 3A = 3^1 + 3^2 + 3^3 + ...+ 3^101
=> 3A-A = 3^101 - 3^0
2A = 3^101 - 1
\(A=\frac{3^{101}-1}{2}\)
b) D = 1 - 5 + 5^2 - 5^3 + ...+ 5^98 - 5^99
=> 5D = 5 - 5^2 + 5^3 - 5^4+...+ 5^99 - 5^100
=> 5D+D = -5^100 + 1
6D = -5^100 + 1
\(D=\frac{-5^{100}+1}{6}\)
a: =2/3+1/5*10/7
=2/3+2/7
=14/21+6/21=20/21
b: \(=\dfrac{1}{2}\cdot\dfrac{-3+2}{4}=\dfrac{1}{2}\cdot\dfrac{-1}{4}=\dfrac{-1}{8}\)
c: \(=\dfrac{3}{4}+\dfrac{9}{5}:\dfrac{3}{2}-1\)
=-1/4+9/5*2/3
=-1/4+18/15
=-1/4+6/5
=-5/20+24/20=19/20
d: \(=\dfrac{3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)
\(=\dfrac{7}{2}-\dfrac{5}{2}\cdot4=\dfrac{7}{2}-\dfrac{20}{2}=\dfrac{-13}{2}\)
a, S = 1 + 2 - 3 - 4 +5 +6 - 7 - 8 +..... +1998 -1999 -2000 +2001
=> S = (1-3)+(2-4)+(5-7)+(6-8)+...+(1997-1999)+... + 2001 ( có 1000 hiệu = -2 )
=> S = -2 x 1000 + 2001 = 1
b, S = 1 - 3 + 5 - 7 + 9 - .... - 1999 + 2001
=> S = (1-3)+(5-7)+(9-11)+....+(1997-1999) + 2001( có 500 hiệu = -2 )
=> S = -2 x 500 + 2001 = 1001
mình chỉ lmf dc 2 câu đầu thông cảm nha
`5/2 -3(1/3-x)=1/4-7x`
`=> 5/2 - 1 + 3x=1/4 -7x`
`=>3x+7x= 1/4 - 5/2 +1`
`=> 10x= 1/4 - 10/4 +4/4`
`=>10x= -5/4`
`=>x=-5/4 :10`
`=>x=-5/4 xx1/10`
`=>x= -5/40=-1/8`
\(A=1+5^2+5^3+...+5^{2015}+5^{2016}\)
\(5A=5+5^3+5^4+...+5^{2016}+5^{2017}\)
\(4A=\left(5+5^3+5^4+...+5^{2016}+5^{2017}\right)-\left(1+5^2+5^3+...+5^{2015}+5^{2016}\right)\)
\(=5+5^{2017}-\left(1+5^2\right)\)
\(=4+5^{2017}-5^2\)
\(A=\frac{4+5^{2017}-5^2}{4}\)
Ta có : 5A = 5 + 5^3 + 5^4 + ... + 5^2016 + 5^2017
=> 5A - A = ( 5 + 5^3 + 5^4 + ... + 5^2016 + 5^2017 ) - ( 1 + 5^2 + 5^3 + ... + 5^2015 + 5^2016 )
=> 4A = 4 + 5^2 + 5^2017
=> A = ( 4 + 5^2 + 5^2017 )/4