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22 tháng 2 2018
\(A=25.3\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
\(=25\left(4-1\right)\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
Áp dụng hằng đẳng thức, ta có : \(A=25\left(4^{1976}-1\right)+25=25.4^{1976}\)
Vậy \(A⋮4^{1976}\)
TT
1
22 tháng 12 2017
đặt B = 42015 + 42014 + 42013 + ... + 42
4B = 42016 + 42015 + 42014 + ... + 43
4B - B = ( 42016 + 42015 + 42014 + ... + 43 ) - ( 42015 + 42014 + 42013 + ... + 42 )
3B = 42016 - 42
\(\Rightarrow\)B = \(\frac{4^{2016}-4^2}{3}\)hay B = \(\frac{4^{2016}-16}{3}\)
\(\Rightarrow\)A = 75 . ( \(\frac{4^{2016}-16}{3}\)+ 5 ) + 25
A = 75 . ( \(\frac{4^{2016}-16}{3}\)+ \(\frac{15}{3}\)) + 25
A = 75 . ( \(\frac{4^{2016}-1}{3}\)) + 25
A = 25 . ( 3 . \(\frac{4^{2016}-1}{3}\)) + 25
A = 25 . ( 42016 - 1 ) + 25
A = 25 . ( 42016 - 1 + 1 )
A = 25 . 42016 \(⋮\)42016
25 tháng 3 2020
a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)
\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)
\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)
nên x-128=0
hay x=128
Vậy: x=128
b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)
\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)
\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)
nên x-1999=0
hay x=1999
Vậy: x=1999
25 tháng 3 2020
a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4
<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)
<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128
b)Tương tự
<=>x-128=0
<=>x=128
Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0
b)tương tự
HH
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