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123456789 , ta co:
2014^9^9^123456789 & 2015^123456789^123456789^9
ta thay 2014 <2015 \ 9<1...9 \9< 1...9
nen 2014^9^9^a < 2015^a^a^9
vi may minh mat vietkey nen khong dang dau thong cam
Ta có
\(A=\frac{2011}{123456789}+\frac{2011}{987654321}+\frac{1}{987654321}\)
\(B=\frac{2011}{123456789}+\frac{1}{123456789}+\frac{2011}{987654321}\)
Mặt khác
\(\frac{1}{987654321}< \frac{1}{123456789}\)
\(\Rightarrow\frac{2011}{123456789}+\frac{2011}{987654321}+\frac{1}{987654321}< \frac{2011}{123456789}+\frac{1}{123456789}+\frac{2011}{987654321}\)
=> A<B
Bài làm:
Ta có:
a) \(x^{n+1}\div5=5^n\)
\(\Leftrightarrow x^{n+1}=5^n.5\)
\(\Leftrightarrow x^{n+1}=5^{n+1}\)
\(\Rightarrow x=5\)
b) \(x^n.9=3^{n+2}\)
\(\Leftrightarrow x^n.3^2=3^{n+2}\)
\(\Leftrightarrow x^n=3^{n+2}\div3^2\)
\(\Leftrightarrow x^n=3^n\)
\(\Rightarrow x=3\)
Ta có:
\(4\left(1+5+5^2+...+5^9\right)=5\left(1+5+5^2+...+5^9\right)-\left(1+5+5^2+...+5^9\right)\)
\(=5+5^2+5^3+...+5^{10}-1-5-5^2-...-5^9\)
\(=5^{10}-1+\left(5-5\right)+\left(5^2-5^5\right)+..+\left(5^9-5^9\right)\)
\(=5^{10}-1\)
=> \(1+5+5^2+...+5^9=\frac{5^{10}-1}{4}\)
Tương tự: \(1+5+5^2+...+5^8=\frac{5^9-1}{4}\)
\(1+3+3^2+...+3^9=\frac{3^{10}-1}{2}\)
\(1+3+3^2+...+3^8=\frac{3^9-1}{2}\)
=> \(A=\frac{5^{10}-1}{5^9-1}>\frac{5^{10}-1}{5^9}=5-\frac{1}{5^9}>4;\)
\(B=\frac{3^{10}-1}{3^9-1}< \frac{3^{10}}{3^9-1}=3+\frac{3}{3^9-1}< 4;\)
=> A > B.
ta co : A = 3/8^3+3/8^4+4/8^4
B=3/8^3+3/8^4+4/8^3
VI 4/8^4 <4/8^3 NEN A<B
Bài 1 : dễ bạn tự làm được :)
Bài 2 :
Ta có :
\(B=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Ta có : B = 2016 + 2017 + 2018 2015 + 2016 + 2017 = 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 Vì : 2016 2015 > 2016 + 2017 + 2018 2015 2017 2016 > 2016 + 2017 + 2018 2016 2018 2017 > 2016 + 2017 + 2018 2017 Nên 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 ⇔ 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 ⇔A > B Vậy A > B Chúc bạn học tốt ~
a,\(\frac{25}{42}-\frac{20}{63}=\frac{5}{18}\)
b,\(\frac{9}{50}-\frac{13}{75}-\frac{1}{6}\)
=\(\frac{1}{150}-\frac{1}{6}\)
=\(-\frac{4}{25}\)
c,\(\frac{2}{15}+\frac{-2}{65}-\frac{4}{39}\)
=\(\frac{32}{195}-\frac{4}{39}\)
=\(\frac{4}{65}\)
a,=75/126-40/126=35/126=5/18
b,=27/150-26/150-25/150=-4/25
c,=26/195+-6/195-20/195=0
2/20x22 + 2/22x24 + 2/24x26 +...+2/78x80
=2 x(1/20x22 + 1/22x24 + 1/24x26 +...+1/78x80)
=2 x(1/20 - 1/22 + 1/22 -....+1/78 - 1/80)
=2 x (1/20 - 1/80)
=2 x 3/80
=3/40
Vì 3/40 < 1/9 nên 2/20x22 + 2/22x24 + 2/24x26 +...+2/78x80 < 1/9
123456789 (có 9 chữ số) < 10^9 (10 chữ số) < 81^9 = 9^18
=> 123456789^9 < 9^(18*9) < 9^(123456789)