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P=(5.311+4.312):(39.52-39.23)
=[311.(5+4.3)]:[39.(52-23)]
=(311.17):(39.17)
=311.17:39:17
=32=9
Vậy P=9
(12+22+32+...+102).(1+3+5...+97+99).(36.333-108.111) [ Đặt (12+22+32+...+102).(1+3+5...+97+99) = A]
= A. (36.3.111-108.111)
= A.(108.111-108.111)
= A.0
= 0
Vậy tích trên bằng 0
(12+22+32+...+102).(1+3+5...+97+99).(36.333-108.111)
= ( 12 + 22 + 32 + ... + 102 ) . ( 1 + 3 +5 + ... + 97 + 99 ) . ( 36 . 3 . 111 - 36 . 3 . 111 )
= ( 12 + 22 + 32 + ... + 102 ) . ( 1 + 3 +5 + ... + 97 + 99 ) . 0
= 0
=\(\left[3^{11}.\left(5+4.3\right)\right]:\left[3^9.\left(5^2-2^3\right)\right]\)
=\(\left[3^{11}.17\right]:\left[3^9.17\right]=\left(3^{11}:3^9\right).17=9.17=153\)
\(\frac{5.3^{11}+4.3^{12}}{3^9.5^2-3^9.2^3}\)
\(=\frac{5.3^{11}+4.3.^{11}}{3^9.\left(5^2-2^3\right)}\)
\(=\frac{5.3^{11}+12.3^{11}}{3^9.\left(25-8\right)}\)
\(=\frac{3^{11}.\left(5+4\right)}{3^9.17}\)
\(=\frac{3^{11}.9}{3^9.17}\)
\(=\frac{3^9.3^2.9}{3^9.17}\)
\(=\frac{3^2.9}{17}\)
\(=\frac{9.9}{17}=\frac{81}{17}\)
Mình chuyển về phân số nhé. k mình nhé
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
a.87 .914 .255 tren 6253 .1812 .243 =(23 )7 .(32 )14 .(52)5 tren (54 )3 .(2.32 )12 .(23 .3)3 =221 .328 .510 tren 512 .(2.32 )12 .(23 .3)3 =217 .325 tren 52 .1.1=217.325 tren 25.
b.(3.128.216 )2 tren (2.4.8.16.32.64)2 =(3.27 .216 )2 tren (2..22 .23 .24 .25 .26 )2 =(3.223 )2 tren (221 )2 =(3.22 )2 tren 1=122 =144
c.4.312 +5.311 tren 39 .23 -39 .52 =4.311 .3+5.311 tren 39 .(23 -52 )=(4+5).311 .3 tren 39 .(8-25)=9.312 tren 39 .(-17)=32 .312 tren 39 .(-17)=314 tren 39 .(-17)=35 tren -17.
hihi ko bt dung ko nua
b/1.2.3.....2013 - 1.2.3....2012-1.2.3...20122
= (1.2.3...2012.2013-1.2.3...2012)-1.2.3...2012.2012
= 1.2.3...2012(2013-1-2012)
=1.2.3...2012.0=0
1.(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^11.3):[3^9.(5^2-2^3)]
=(5.3^11+12.3^11):[3^9.17]
=3^11.(5+12):(3^9.17)
=(3^11.17):(3^9.17)
=17.(3^11:3^9)
=17.3^2
=17.9
=153
2.(12+22+32+...+992+1002).(36.333-108.111)
=2.(12+22+32+...+992+1002).(36.333-36.3.111)
=2.(12+22+32+...+992+1002).(36.333-36.333)
=2.(12+22+32+...+992+1002).0
=0