=(99.100.101-0.1.2):3=333300

Đặt A= 1.2+2.3 +.......+99.100

3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3

3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)

3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)

3A = 99.100.101 - 0.1.2

3A = 999900 - 0

3A= 999900

A= 999900 : 3

A = 333300

Học tốt nha!

Đặt A = 1.2 + 2.3 + ... + 99.100

=> 3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100

=> 3A = 99.100.101

=> A = 333300

140.28

bạn giải ra đi

3s1=(2.3+3.4+4.5+5.6+...+15.16+16.17).3

3s1=2.3.3+3.4.3+4.5.3+5.6.3+....+15.16.3+16.17.3

3s1=2.3(4-1)+3.4(5-2)+4.5(6-3)+5.6(7-4)+...+15.16(17-14)+16.17(18-15)

3s1=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+15.16.17-14.15.16+16.17.18-15.16.17

3s1=(16.17.18-1.2.3)/3

s1=1630

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{6}{21}=\dfrac{2}{7}\)

\(\dfrac{2}{7}\)

-19,35

\(=-3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)=-3\cdot\dfrac{5}{14}=-\dfrac{15}{14}\)

tính binh thường trong mày tính thì sẽ bằng \(\dfrac{207}{20}\)

6,45

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\\ =\dfrac{1}{2}-\dfrac{1}{7}\\ =\dfrac{5}{14}\)

= 3.(3/3 - 3/4 + 3/4 - 3/5 + 3/5 - 3/6 +.....+ 3/277 - 3/278 + 3/278 - 3/279)

= 3.(3/3 - 3/279 )

= 3.92/93

= 187/93

Mình cũg không chắc chắn là 100% đâu bạn nên dò lại nhé

Đặt biểu thức là \(A\)

\(A=\dfrac{3}{3.4}+\dfrac{3}{4.5}+\dfrac{3}{5.6}+...+\dfrac{3}{278.279}\)

\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}\left(\dfrac{3}{3.4}+\dfrac{3}{4.5}+\dfrac{3}{5.6}+...+\dfrac{3}{278.279}\right)\)

\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{278.279}\)

\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{278}-\dfrac{1}{279}\)

\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3}-\dfrac{1}{279}\)

\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{93}{279}-\dfrac{1}{279}\)

\(\Leftrightarrow\dfrac{1}{3}A=\dfrac{92}{279}\)

\(\Leftrightarrow A=\dfrac{92}{279}:\dfrac{1}{3}\)

\(\Leftrightarrow A=\dfrac{92}{279}.3\)

\(\Leftrightarrow A=\dfrac{92}{93}\)