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DP
21 tháng 7 2017
\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{16}\right)+\left(1+\frac{1}{32}\right)+\left(1+\frac{1}{64}\right)-7\)
\(S=\left(1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-7\)
\(S=6+\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+....+\left(\frac{1}{32}-\frac{1}{64}\right)\right]-7\)
\(S=6+\left(1-\frac{1}{64}\right)-7\)
\(S=6+\frac{63}{64}-7\)
\(S=\frac{447}{64}-7=-\frac{1}{64}\)
19 tháng 6 2016
\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-1\)
\(S+1=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(2\left(S+1\right)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(2\left(S+1\right)-\left(S+1\right)=S+1=1-\frac{1}{2^6}=\frac{63}{64}\)
\(S=\frac{63}{64}-1\)
MH
12 tháng 6 2015
a.\(\frac{3}{5}\).x-( -\(\frac{1}{2}\))2=\(\sqrt{\frac{9}{16}}\)
<=> \(\frac{3.x}{5}\)-\(\frac{1}{4}\)=\(\frac{3}{4}\)
<=> \(\frac{3.x}{5}\)=1
<=> 3x=5
<=> x= \(\frac{5}{3}\)
b, x-\(\frac{1}{2}\)+\(\frac{1}{3}\)=\(2\frac{1}{3}\)
<=> x=\(2\frac{1}{3}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)
<=> x=2\(\frac{1}{2}\)
c,33x: 11x =81
<=> ( 3.11)x: 11x=81
<=> 3x. 11x; 11x=81
<=> 3x=81
<=> 3x= 34
<=> x= 4
12 tháng 6 2015
a, \(\frac{3}{5}x-\frac{1}{4}=\frac{3}{4}\)
\(\frac{3}{5}x=\frac{3}{4}+\frac{1}{4}\)
\(\frac{3}{5}x=1\)
\(x=1:\frac{3}{5}\)
\(x=\frac{5}{3}\)
b, \(x-\frac{1}{2}+\frac{1}{3}=\frac{7}{3}\)
\(x=\frac{7}{3}-\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{2}\)
c, ( 33:11 )x = 81
3x = 81
3x = 34
=> x = 4
Nhớ tích đúng cho mh nha !!!!!!!!!!!
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(\begin{array}{l}0,49 = {\left( {0,7} \right)^2};\\\,\frac{1}{{32}} =\frac{1^5}{2^5}={\left( {\frac{1}{2}} \right)^5};\\\,\frac{{ - 8}}{{125}} =\frac{(-2)^3}{5^3}= {\left( {\frac{{ - 2}}{5}} \right)^3};\end{array}\)
\(\frac{{16}}{{81}} =\frac{4^2}{9^2}= {\left( {\frac{4}{9}} \right)^2} (hoặc \,\frac{{16}}{{81}} =\frac{2^4}{3^4}= {\left( {\frac{2}{3}} \right)^4});\\\,\frac{{121}}{{169}} =\frac{11^2}{13^2}= {\left( {\frac{{11}}{{13}}} \right)^2}\)
TT
1
2 tháng 8 2018
\(30A=\frac{30^{32}+30}{30^{32}+1}=\frac{30^{32}+1+29}{30^{32}+1}=1+\frac{29}{30^{32}+1}\)
\(30B=\frac{30^{33}+30}{30^{33}+1}=\frac{30^{33}+1+29}{30^{33}+1}=1+\frac{29}{30^{33}+1}\)
Vì \(\frac{29}{30^{32}+1}>\frac{29}{30^{33}+1}\) nên \(1+\frac{29}{30^{32}+1}>1+\frac{29}{30^{33}+1}\Rightarrow30A>30B\Rightarrow A>B\)
Vậy \(A>B.\)
Chúc bạn học tốt.
Ta có: \(9.3^3.\frac{1}{81}.3^2\)
\(=3^2.3^3.\frac{1}{3^4}.3^2\)
\(=3^3=27\)
Bài làm :
Ta có :
\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\frac{1}{3^4}.3^2=\frac{3^7}{3^4}=3^3=27\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!