b, S = 4 x 4 x 4 x...x 4 (2013 số 4)
c, S = 9 x 9 x 9 x....x 9 (2002 số 9)
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AH
Akai Haruma
Giáo viên
19 tháng 12 2017
Lời giải:
\(a+b=3\Rightarrow a+(b-2)=1\Rightarrow b-2=1-a\)
Ta có:
\(f(x)=\frac{9^x}{9^x+3}\Rightarrow f(a)=\frac{9^a}{9^a+3}\) (1)
\(f(b-2)=f(1-a)=\frac{9^{1-a}}{9^{1-a}+3}=\frac{9}{9^a\left(\frac{9}{9^a}+3\right)}\)
\(=\frac{9}{9+3.9^a}=\frac{3}{3+9^a}\) (2)
Từ (1),(2) suy ra \(f(a)+f(b-2)=\frac{9^a}{9^a+3}+\frac{3}{3+9^a}=\frac{9^a+3}{9^a+3}=1\)
Đáp án A
9 tháng 7 2018
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
T
1
17 tháng 8 2021
Đáp án :
\(\Rightarrow\)= 92002 \(\Rightarrow\)chữ số tận cùng là 1
# Hok tốt !
Y
23 tháng 7 2018
a) \(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)
b) \(\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{64}-1\right)\)
\(=\dfrac{9^{64}-1}{10}\)
Ta có: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)
Mà \(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)
\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)
c) Ta có:
\(\dfrac{x^2-y^2}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-xy}\left(1\right)\)
Vì x>y>0, ta có:
\(\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)
Vì x>y>0 nên \(\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)\)
Từ (1), (2) và (3) suy ra:
\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)
PT
18 tháng 9 2018
a) Ta có:
\(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
Vậy 2011.2013+2012.2014 = 20122 + 20132 - 2
GP
1
VN
1 tháng 7 2018
a, -3/4 + 3/7 + -1/4 + 4/9 + 4/7
=\(\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
=\(-1+1+\frac{4}{9}\)
=\(\frac{4}{9}\)
b, -7/9 x 4/11 + -7/9 x 7/11 + 5 và 7/9
\(=\frac{7}{9}x\frac{-4}{11}+\frac{7}{9}x\frac{-7}{11}+5x\frac{7}{9}\)
\(=\frac{7}{9}x\left(\frac{-4}{11}+\frac{-7}{11}+5\right)\)
\(=\frac{7}{9}x4\)
\(=\frac{28}{9}\)
c, ( 21/31 + 2013/6039 ) - ( 44/53 - 10/31) - /53
4 tháng 4 2018
a) 2 x 2 x 2 x ... x 2 x 2 :2
b) 3 x 3 x 3 x ... x 3 x 3 :3
c) 4 x 4 x 4 x ... x 4 x 4 :4
d) 5 x 5 x 5 x ... x 5 x 5 :5
e) 6 x 6 x 6 x ... x 6 x 6 :6
g) 7 x 7 x 7 x...x 7 x 7 :7
h) 8 x 8 x 8 x... x 8 x 8 :8
i) 9 x 9 x 9 x...x 9 x 9 :9
giúp mik đi đúng mik tick cho :(
\(b,S=4\times4\times4\times...\times4\\ S=4^{2013}\\ S=2^{4026}\\ c,S=9\times9\times9\times...\times9\\ \Rightarrow S=9^{2002}\)