1/2+1/3+14

= 5/6 + 14

= 89/6

Chúc bạn học tốt nhé! ^-^

=5/6+14

=89/6

(+) \(S=1+2+3+...+200\)

\(\Rightarrow S=\frac{\left(200+1\right)200}{2}=20100\)

(+) \(A=1.2+2.3+....+199.200\)

\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+.....+199.200.\left(201-198\right)\)

\(\Rightarrow3A=1.2.3-0.1.2+2.3.4-1.2.3+.....+199.200.201-198.199.200\)

\(\Rightarrow3A=199.200.201\)

\(\Rightarrow A=\frac{199.200.201}{3}\)

\(\Rightarrow A=2666600\)

(+) \(B=1^2+2^2+....+200^2\)

\(\Rightarrow B=1\left(0+1\right)+2\left(1+1\right)+3\left(2+1\right)+....+200\left(199+1\right)\)

\(\Rightarrow B=1+1.2+2+2.3+3+.....+199.200+200\)

\(\Rightarrow B=\left(1+2+3+....+200\right)+\left(1.2+2.3+....+199.200\right)\)

\(\Rightarrow B=S+A\)

\(\Rightarrow B=2666600+20100=2686700\)

S = 1 + 2 + 3 + ... + 200

S = (1 + 200).200:2

S = 201.100

S = 20100

A = 1.2 + 2.3 + ... + 199.200

3A = 1.2.(3-0) + 2.3.(4-1) + ... + 199.200.(201-198)

3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + ... + 199.200.201 - 198.199.200

3A = 199.200.201

A = 199.200.67

A = 2666600

B = 1² + 2² + 3² + ... + 200²

B = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + ... + 200.(199 + 1)

B = 0.1 + 1.1 + 1.2 + 1.2 + 2.3 + 1.3 + ... + 199.200 + 1.200

B = (0.1 + 1.2 + 2.3 + ... + 199.200) + (1.1 + 1.2 + 1.3 + ... + 1.200)

B = 2666600 + (200 + 1).200:2

B = 2666600 + 201.100

B = 2666600 + 20100

B = 2686700

\(\frac{1}{2}\div\frac{3}{4}+\frac{1}{6}\div\frac{3}{4}\)

\(=\frac{3}{4}\div\left(\frac{1}{2}+\frac{1}{6}\right)\)

\(=\frac{3}{4}\div\frac{2}{3}\)

\(=\frac{9}{8}\)

\(\frac{1}{2}\div\frac{3}{4}-\frac{1}{6}\div\frac{3}{4}\)

\(=\left(\frac{1}{2}-\frac{1}{6}\right)\div\frac{3}{4}\)

\(=\frac{1}{3}\div\frac{3}{4}\)\

\(=\frac{4}{9}\)

1/2 : 3/4 + 1/6 : 3/4 = 1/2 x 4/3 + 1/6 x 4/3 = 4/3 x (1/2 + 1/6) = 4/3 x 2/3 = 8/9

1/2 : 3/4 - 1/6 : 3/4 = 1/2 x 4/3 - 1/6 x 4/3 = 4/3 x (1/2 - 1/6) = 4/3 x 1/3 = 4/9

Ta đã biết \(\dfrac{1}{a\cdot a}< \dfrac{1}{\left(a+1\right)\left(a-1\right)}\) ( a ϵ Z )

⇒ \(Q=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{200\cdot200}\) < \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\) 

Ta có \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\) 

= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)

= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)

= \(\dfrac{1}{2}\left(1-\dfrac{1}{201}\right)=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}< \dfrac{100}{200}=\dfrac{1}{2}< \dfrac{3}{4}\)

Vậy Q < \(\dfrac{3}{4}\)

\(\frac{2}{3}x=\frac{5}{4}+\frac{1}{2}x\)

\(\Rightarrow\frac{2}{3}x-\frac{1}{2}x=\frac{5}{4}\)

\(\frac{1}{6}x=\frac{5}{4}\)

\(x=\frac{5}{4}:\frac{1}{6}\)

x = 15/2

2/3.x = 5/4 + 1/2 . x

2/3. x - 1/2 . x =5/4

(2/3-1/2).x=5/4

1/3.x=5/4

x=5/4:1/3

x=15/4