Công thức tìm số bé:(T-H):2
⇒Số bé:40:2=20
      Đáp số:20

40:2=20

đáp số : 20

\(x+\dfrac{1}{4}=\dfrac{7}{2}\)

\(x=\dfrac{7}{2}-\dfrac{1}{4}\)

\(x=\dfrac{14}{4}-\dfrac{1}{4}\)

\(x=\dfrac{13}{4}\)

\(x\)  + \(\dfrac{1}{4}\) = \(\dfrac{7}{2}\)

\(x\)         = \(\dfrac{7}{2}\) - \(\dfrac{1}{4}\)

\(x\)        = \(\dfrac{13}{4}\)

a: D

b: D

a)1/2;3/4;7/8;15/16

B)20/2;27/20;45/40;21/20

a: \(\dfrac{3}{4}=\dfrac{3\cdot4}{4\cdot4}=\dfrac{12}{16};\dfrac{1}{2}=\dfrac{1\cdot8}{2\cdot8}=\dfrac{8}{16};\dfrac{7}{8}=\dfrac{7\cdot2}{8\cdot2}=\dfrac{14}{16}\)

mà 8<12<14<15

nên \(\dfrac{1}{2}< \dfrac{3}{4}< \dfrac{7}{8}< \dfrac{15}{16}\)

b: 20/2=10; 21/20=1,05; 45/40=1,125; 27/20=1,35

mà 10>1,35>1,125>1,05

nên \(\dfrac{20}{2}>\dfrac{27}{20}>\dfrac{45}{40}>\dfrac{21}{20}\)

\(\dfrac{3}{5}=\dfrac{3\cdot3}{5\cdot3}=\dfrac{9}{15}\)

\(\dfrac{13}{15}=\dfrac{13\cdot1}{15\cdot1}=\dfrac{13}{15}\)

mà 9<13

nên \(\dfrac{3}{5}< \dfrac{13}{15}\)

\(\dfrac{3}{10}=\dfrac{3\cdot1}{10\cdot1}=\dfrac{3}{10};\dfrac{2}{5}=\dfrac{2\cdot2}{5\cdot2}=\dfrac{4}{10}\)

mà 3<10

nên \(\dfrac{3}{10}< \dfrac{2}{5}\)

Vì 19<17

nên \(\dfrac{13}{19}>\dfrac{13}{17}\)

\(\dfrac{3}{5}>\dfrac{13}{15}\\ \dfrac{3}{10}< \dfrac{2}{5}\\ \dfrac{13}{19}< \dfrac{13}{17}.\)

M=1x2+2x3+3x4+xxx+19x20

3M=1x2x3+2x3x3+3x4x3+xxx+19x20x3

3M=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+xxx+19x20x(21-18)

3M=(1x2x3-0x1x2)+(2x3x4-1x2x3)+(3x4x5-2x3x4)+xxx+(19x20x21-18x19x20)

3M=19x20x21-0x1x2

3M=7980-0

3M=7980

M=7980:3

M=2660

Vậy M=2660

Chúc bạn học tốt nha!!!!! ^^

\(A=1\cdot2+2\cdot3+...+19\cdot20\)

\(=1\left(1+1\right)+2\left(2+1\right)+...+19\left(19+1\right)\)

\(=\left(1^2+2^2+...+19^2\right)+\left(1+2+...+19\right)\)

\(=\dfrac{19\left(19+1\right)\left(2\cdot19+1\right)}{6}+\dfrac{19\cdot20}{2}\)

\(=2470+190=2660\)

504 nha

M xin cách làm với

\(\dfrac{6}{56}=\dfrac{6:2}{56:2}=\dfrac{3}{28}\)