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a, 5/9-1/6+1/3 =\(\dfrac{13}{18}\) b, 1/8:9/12+1/3=\(\dfrac{1}{2}\)
c, 5/4x4/3x3/4 = \(\dfrac{5}{4}\) d, 8/5x(3/7+1/4) =\(\dfrac{38}{35}\)
a)10/18-3/18+6/18
=13/18
b)1/8.12/9+1/3
=1/6+1/3
=1/6+2/6
=1/2
c)5.4.3/4.3.4=5/4
d)8/5.(12/28+7/28)
=8/5.19/28=38/35
Chứng minh:
C=\(\dfrac{1}{2x2}\)+\(\dfrac{1}{3x3}\)+\(\dfrac{1}{4x4}\)+.....+\(\dfrac{1}{100x100}\)<1
\(C=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{100\times100}\\ C< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ C< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ C< 1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
= 385
1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10
= 1+4+9+16+25+36+49+64+81+100
=(81+9)+(64+16)+(49+1)+)36+4)+25+100
=90+80+50+40+25 +100
=385
\(E=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}\)
Ta có \(\frac{1}{2.2}>\frac{1}{2.3}\)
\(\frac{1}{3.3}>\frac{1}{3.4}\)
...
\(\frac{1}{49.49}>\frac{1}{49.50}\)
=> \(E=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{49.49}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}=F\)
=> E > F
2/3 . 3/4 . 4/5
bằng 2/5