Bài 7:

Số phần kẹo Hùng đã cho Hà và Hồng là:

\(\dfrac{2}{7}+\dfrac{1}{7}=\dfrac{3}{7}\left(phần\right)\)

Hùng còn lại số phần của gói kẹo là:

\(\dfrac{6}{7}-\dfrac{3}{7}=\dfrac{3}{7}\left(phần\right)\)

1:

 2 3/4

5 6/5

3 3/9

7 6/8

2:

1/3 + 2/3 + (3/4 + 1/4) = 2

=2

= 4 5/10

1:

\(2+\dfrac{3}{4}=\dfrac{8}{4}+\dfrac{3}{4}=\dfrac{11}{4};5+\dfrac{6}{5}=\dfrac{25}{5}+\dfrac{6}{5}=\dfrac{31}{5};\dfrac{3}{9}+3=\dfrac{1}{3}+\dfrac{9}{3}=\dfrac{10}{3}\)

2: 

a: \(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\\ =1+1=2\)

b: \(=\left(\dfrac{3}{4}+\dfrac{2}{8}\right)+\left(\dfrac{3}{5}+\dfrac{4}{10}\right)\\ =\left(\dfrac{6}{8}+\dfrac{2}{8}\right)+\left(\dfrac{6}{10}+\dfrac{4}{10}\right)\\ =1+1=2\)

c:\(=\left(\dfrac{1}{10}+\dfrac{9}{10}\right)+\left(\dfrac{2}{10}+\dfrac{8}{10}\right)+\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{4}{10}+\dfrac{6}{10}\right)+\dfrac{5}{10}\\ =1+1+1+1+\dfrac{5}{10}=4\dfrac{5}{10}\)

\(\frac{9}{10}\)+\(\frac{7}{9}\)+\(\frac{5}{8}\)+\(\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)

\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)\)\(+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)

\(=1+1+1+1+1\)

\(=5\)

9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10

= ( 9/10 + 1/10 ) + ( 7/9 + 2/9 ) + ( 5/8 + 3/8 ) + ( 3/7 + 4/7 ) + ( 3/5 + 2/5 )

= 1 + 1 + 1 + 1 + 1

= 5

= (9/10+ 1/10 )+(7/9 + 2/9)+( 5/8 +3/8)+(3/5 +2/5)

=1+1+1+1=4

Gọi biểu thức trên là A, ta có :

A       = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 9x10

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 9x10x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 9x10x(11-9)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 9x10x11 - 8x9x10.

A x 3 = 9x10x11

A       = 9x10x11 : 3

A       =330

= 331 nha !

    CHÚC NGÀY 1 THÁNG 6 VUI VẺ 

330 mới đúng

S = 1 + 2 + 2² + ... + 2⁹ + 2¹⁰

2S = 2 . ( 1 + 2 + 2² + ... + 2⁹ + 2¹⁰ )

2S = 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹

2S - S = ( 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹ ) - ( 1 + 2 + 2² + ... + 2⁹ + 2^{10 )}

S = 2¹¹ - 1

S = 1 + 2 + 2² + ... + 2⁹ + 2¹⁰

2S = 2 . ( 1 + 2 + 2² + ... + 2⁹ + 2¹⁰ )

2S = 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹

2S - S = ( 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹ ) - ( 1 + 2 + 2² + ... + 2⁹ + 2^{10 )}

S = 2¹¹ - 1

S = 1 + 2 + 2² + ... + 2⁹ + 2¹⁰

2S = 2 . ( 1 + 2 + 2² + ... + 2⁹ + 2¹⁰ )

2S = 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹

2S - S = ( 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹ ) - ( 1 + 2 + 2² + ... + 2⁹ + 2^{10 )}

S = 2¹¹ - 1

\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)

= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)

= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)

= \(1+1+1+1+1\)

= \(1\times5\)

= \(5\)

Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :

A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10

A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)

A = 1 + 1 + 1 + 1 + 1

A = 5

Ta có \(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)

\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{3}{8}+\frac{5}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\)

\(=1+1+1+1+1\)

\(=5\)