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Ta có : A=20/11×13 + 20/13×15 +20/15×17+...+20/53×55
A = 10 ×( 2/11×13+2/13×15+...12/53×55)
A = 10 ×(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)
A = 10 × (1/11-1/55)
A =10 × 4/55
A = 8/11
đầu bài: 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20
= (1+19)+(2+18)+(3+17)+(4+16)+(5+15)+(6+14)+(7+13)+(8+12)+(9+11)+20
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 +20
= 20.10
= 200
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=\(\frac{\left(1+20\right)20}{2}\)=210
Bài 1:
a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)
\(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{3}\)
\(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)
\(\dfrac{1}{4}\) \(x\)= \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)
\(\dfrac{1}{4}x=\dfrac{11}{24}\)
\(x=\dfrac{11}{24}:\dfrac{1}{4}\)
\(x=\dfrac{11}{24}\times4\)
\(x=\dfrac{11}{6}\)
b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)
\(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)
\(x\) = \(\dfrac{9}{10}\)
Ta có:
A = 1 + 3 + 5 + 7 +... + 101
A = \(\frac{102.51}{2}=2601\)
M = 16 - 18 + 20 - 22 + 24 - 26 + .. + 64 - 66 + 68
M = ( 16 - 18 ) + ( 20 - 22 ) + ( 24 - 26 ) + ... + ( 64 - 66 ) + 68
M = (- 2 + - 2 + -2 + ... + - 2 ) + 68
M = 25/2 . ( - 2 ) + 68
M = -25 + 68
M = 43
H = ( 1 + 2 + 3 +...+ 99 ) x ( 13 x 15 - 12 x 15 - 15 )
H = ( 1 + 2 + 3 +...+ 99 ) x { (13 - 12 - 1) x 15 }
H = ( 1 + 2 + 3 +...+ 99 ) x 0
H = 0
G = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + 13 + 14 - ... + 301 + 302
G = ( 1 + 2 ) + ( -3 - 4 ) + ( 5 + 6 ) + ( -7 - 8 ) + ( 9 + 10 ) + ( - 11 - 12 ) + ( 13 + 14 ) -...+ ( 301 + 302 )
G = ( 3 - 7 ) + ( 11 - 15 ) + ( 19 - 23 ) + 27 - ... + 603
G = -4 + - 4 + -4 + 27 - ... + 603
G = 75 x ( -4 ) + 603
G = -300 + 603
G = 303
2.
a) 1 + 2 + 3 + 4 +...+ 99 + 100 + 2 x X = 5052
= > \(\frac{100.101}{2}\)+ 2 x X = 5052
= > 5050 + 2 x X = 5052
= > 2X = 2
= > X = 1
a) \(\frac{A}{7}=\frac{5}{2\times7}+\frac{4}{7\times11}+\frac{3}{11\times14}+\frac{1}{14\times15}+\frac{13}{15\times28}\)
\(\frac{A}{7}=\frac{7-2}{2\times7}+\frac{11-4}{7\times11}+\frac{14-11}{11\times14}+\frac{15-14}{14\times15}+\frac{28-15}{15\times28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}=\frac{13}{4}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{100}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{100}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{49}{100}\)
<=> \(\frac{1}{x+1}=\frac{1}{100}\)
<=> x + 1 = 100
<=> x = 99
b) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
<=> \(x-10.\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
<=> \(x-10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
<=> \(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
<=> \(x-\frac{8}{11}=\frac{3}{11}\)
<=> x = 1
Vậy x = 1
P/S : Dấu "." là dấu "x"