Đặt \(A=\left(1-\dfrac{2}{42}\right)\left(1-\dfrac{2}{56}\right)\left(1-\dfrac{2}{72}\right)...\left(1-\dfrac{2}{2652}\right)\)

\(=\left(1-\dfrac{2}{6.7}\right)\left(1-\dfrac{2}{7.8}\right)\left(1-\dfrac{2}{8.9}\right)...\left(1-\dfrac{2}{51.52}\right)\)

Ta có:

\(1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Do đó:

\(A=\dfrac{5.8}{6.7}.\dfrac{6.9}{7.8}.\dfrac{7.10}{8.9}...\dfrac{50.53}{51.52}\)

\(=\dfrac{5.6.7...50}{6.7.8...51}.\dfrac{8.9.10...53}{7.8.9...52}=\dfrac{5}{51}.\dfrac{53}{7}=\dfrac{265}{357}\)

Ta có: GH//JI

=>\(\widehat{JGH}+\widehat{GJI}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{JGH}=180^0-90^0=90^0\)

ta có: GH//JI

=>\(\widehat{HIJ}=\widehat{xHI}\)(hai góc so le trong)

=>\(\widehat{HIJ}=47^0\)

Sửa đề: 

`S = 1/3 + 2/(3^2) + 3/(3^3) + ... + 100/(3^100)`

`3S = 1 + 2/3 + 3/(3^2) + ... + 100/(3^99)`

`3S - S = 1 - 100/3^100 + (2/3 - 1/3) + (3/(3^2) - 2/(3^2)) + ... + (100/(3^99) - 99/(3^99)) `

`2S = 1 - 100/(3^100) + 1/3 + 1/(3^2) + ... + 1/(3^99) `

Đặt `A = 1/3 + 1/(3^2) + ... + 1/(3^99) `

`=> 3A = 1 + 1/3 + ... + 1/(3^98) `

`=> 3A - A = (1 + 1/3 + ... + 1/(3^98)) - ( 1/3 + 1/(3^2) + ... + 1/(3^99) )`

`=> 2A = 1 - 1/(3^99)`

`=> A = (1 - 1/(3^99))/2`

Khi đó: `2S = 1 - 100/(3^100) + (1 - 1/(3^99))/2`

`S = 1/2 - 100/(2.3^100) + (1 - 1/(3^99))/4`

Ta có: `{(1/2 - 100/(2.3^100) < 1/2),((1 - 1/(3^99))/4 < 1/4):}`

`=>  1/2 - 100/(2.3^100) + (1 - 1/(3^99))/4 < 1/2 + 1/4 = 3/4`

Hay `S < 3/4 (đpcm)`

3|4lớn hơn

tk nhé

Ông An cao 180 cm, vòng bụng 108 cm.

Ông Chung cao 160 cm, vòng bụng 70 cm.

a: |2,5|+|7,5|=2,5+7,5=10

b: \(1,2\cdot\left|-3\right|+6,4=1,2\cdot3+6,4=3,6+6,4=10\)

c: \(\left|-\dfrac{7}{2}\right|+\left|\dfrac{15}{2}\right|=\dfrac{7}{2}+\dfrac{15}{2}=\dfrac{22}{2}=11\)

Do 8 chia hết cho 4 \(\Rightarrow8^{2008}⋮4\)

\(\Rightarrow8^{2008}=4k\)

\(\Rightarrow5^{8^{2008}}=5^{4k}=\left(5^4\right)^k=625^k\)

Mà \(625\equiv1\left(mod24\right)\Rightarrow625^k\equiv1\left(mod24\right)\)

\(\Rightarrow5^{8^{2008}}\equiv1\left(mod24\right)\)

\(\Rightarrow5^{8^{2008}}+23\equiv0\left(mod24\right)\)

Hay \(5^{8^{2008}}+23\) chia hết 24

a.

Do \(My||BC\Rightarrow\widehat{CMy}=\widehat{MCB}\) (so le trong)

Mà \(\widehat{MCB}=45^0\Rightarrow\widehat{CMy}=45^0\)

lại có My là phân giác của \(\widehat{CMx}\Rightarrow\widehat{CMx}=2\widehat{CMy}\)

\(\Rightarrow\widehat{CMx}=2.45^0=90^0\)

b.

Do \(BC||My\Rightarrow\widehat{CBM}=\widehat{xMy}\)

Mà \(\widehat{xMy}=\widehat{CMy}=45^0\) (My là phân giác)

\(\Rightarrow\widehat{CBM}=45^0\)

Lại có Bx là phân giác \(\widehat{ABC}\Rightarrow\widehat{ABC}=2\widehat{CBM}\)

\(\Rightarrow\widehat{ABC}=2.45^0=90^0\)

\(\Rightarrow\Delta ABC\) vuông tại B

Áp dụng công thức: \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(\Rightarrow1-\dfrac{1}{1+2+...+n}=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=1-\dfrac{2}{n\left(n+1\right)}\)

\(=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Do đó:

\(A=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{4.5.6...\left(n+2\right)}{3.4.5...\left(n+1\right)}=\dfrac{1}{n}.\dfrac{n+2}{3}=\dfrac{n+2}{3n}\)

\(\Rightarrow A=\dfrac{B}{3}\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{3}\)

a: m\(\perp\)a

n\(\perp\)a

Do đó: m//n

b: m//n

=>\(\widehat{A_1}=\widehat{ABC}\)(hai góc so le trong)

=>\(\widehat{A_1}=72^0\)

c: Xét ΔABC có \(\widehat{BAC}+\widehat{ACB}+\widehat{ABC}=180^0\)

=>\(\widehat{C_1}=180^0-64^0-72^0=44^0\)

\(A=\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

\(299A=1+\dfrac{1}{2}+...+\dfrac{1}{101}-\left(\dfrac{1}{300}+\dfrac{1}{301}+...+\dfrac{1}{400}\right)\)

Thêm bớt \(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{299}\) ta được:

\(299A=1+\dfrac{1}{2}+...+\dfrac{1}{101}+\left(\dfrac{1}{102}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{300}+...+\dfrac{1}{400}\right)\)

\(299A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{400}\right)\)

\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+....+\dfrac{1}{299}-\dfrac{1}{400}\)

\(101B=\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{400}\right)\)

\(\Rightarrow299A=101B\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{101}{299}\)