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Câu 1:
0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3
= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059
= 9 x 43,6 + 9 x 85,8 + 6 x 105,9
= 3 x 130,8 + 3 x 257,4 + 3 x 211,8
= 3 x ( 130,8 + 257,4 + 211,8 )
= 3 x 600
= 1800
Câu 2:
3/4 x X + 1/2 x X - 15 = 35
X x ( 3/4 + 1/2 ) - 15 = 35
X x ( 3/4 + 1/2 ) = 50
X x 5/4 = 50
X = 40
VẬy X = 40
\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)
Dấu \(.\)là dấu nhân
Ta có :
\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(\Rightarrow E=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{2}{100.103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\frac{102}{103}\)
\(\Rightarrow E=\frac{68}{103}\)
Vậy \(E=\frac{68}{103}\)
~ Ủng hộ nhé
\(E=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+...+\frac{2}{100\cdot103}\)
\(E=2\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}\right)\)
Gọi tổng trong ngoặc là F
\(\Rightarrow3F=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\)
\(\Rightarrow3F=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(\Rightarrow3F=1-\frac{1}{103}=\frac{102}{103}\)
\(\Rightarrow F=\frac{102}{103\cdot3}=\frac{34}{103}\)
\(\Leftrightarrow E=2\cdot\frac{34}{103}=\frac{68}{103}\)
Vậy......
Đặt \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\frac{102}{103}\)
\(\Rightarrow B=\frac{68}{103}\)
Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\cdot\frac{102}{103}\)
\(A=\frac{68}{103}\)
\(\frac{2}{1\times4}+\frac{2}{4\times7}+\frac{2}{7\times10}+...+\frac{2}{37\times40}\)
\(=\frac{2}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{37\times40}\right)\)
\(=\frac{2}{3}\times\left(\frac{4-1}{1\times4}+\frac{7-4}{4\times7}+\frac{10-7}{7\times10}+...+\frac{40-37}{37\times40}\right)\)
\(=\frac{2}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)
\(=\frac{2}{3}\times\left(1-\frac{1}{40}\right)=\frac{13}{20}\)
1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
\(A=\dfrac{5}{4x7}+\dfrac{5}{7x10}+...+\dfrac{5}{25x28}+\dfrac{5}{28x31}\)
\(\dfrac{3}{5}A=\dfrac{7-4}{4x7}+\dfrac{10-7}{7x10}+...+\dfrac{28-25}{25x28}+\dfrac{31-28}{28x31}\)
\(\dfrac{3}{5}A=\dfrac{7}{4x7}-\dfrac{4}{4x7}+\dfrac{10}{7x10}-\dfrac{7}{7x10}+...+\dfrac{28}{25x28}-\dfrac{25}{25x28}+\dfrac{31}{28x31}-\dfrac{28}{28x31}\)
\(\dfrac{3}{5}A=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}+\dfrac{1}{28}-\dfrac{1}{31}\)
\(\dfrac{3}{5}A=\dfrac{1}{4}-\dfrac{1}{31}=\dfrac{27}{124}\)
\(A=\dfrac{27}{124}:\dfrac{3}{5}=\dfrac{27}{124}x\dfrac{5}{3}=\dfrac{45}{124}\)
\(\left(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{3001.3004}\right)\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{3001}-\frac{1}{3004}\right)\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{2}{3}\cdot\left(1-\frac{1}{3004}\right)\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{2}{3}\cdot\frac{3003}{3004}\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow\frac{1001}{1502}\cdot\left(x+1\right)=\frac{9009}{1502}\)
\(\Leftrightarrow x+1=\frac{9009}{1502}\div\frac{1001}{1502}\)
\(\Leftrightarrow x+1=9\Rightarrow x=8\)