a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)

\(=\dfrac{3.4.5...100}{2.3.4...99}\)

\(=\dfrac{100}{2}=50\)

a,

\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)

b,

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)

c,

\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)

b.\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\)

\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-99}{100}\)

\(=\dfrac{-1.\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}\)

\(=-\dfrac{1}{100}\)

= 3/10 + 4/7 . 5/4 - 1/10 
= 3/10 + 5/7 - 1/70 
= 1.

\(\dfrac{3}{10}+\dfrac{4}{7}.\dfrac{5}{4}-\dfrac{1}{70}=\dfrac{3}{10}+\dfrac{5}{7}-\dfrac{1}{70}=\dfrac{21}{70}+\dfrac{50}{70}-\dfrac{10}{70}=\dfrac{70}{70}=1\)

1) \(=\dfrac{1}{2}.\dfrac{3}{5}+\dfrac{4}{7}.\dfrac{5}{4}-\dfrac{1}{70}\)

\(=\dfrac{3}{10}+\dfrac{5}{7}-\dfrac{1}{70}\)

\(=\dfrac{21}{70}+\dfrac{50}{70}-\dfrac{1}{70}\)

\(=1\)

2)

\(=\left(\dfrac{30}{45}+\dfrac{9}{45}-\dfrac{20}{45}\right):\left(\dfrac{15}{45}+\dfrac{18}{45}-\dfrac{30}{45}\right)\)

\(=\dfrac{19}{45}:\dfrac{1}{45}\)

\(=\dfrac{19}{45}.45\)

\(=19\)

\(S=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{30}}\)

\(\Rightarrow4S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{29}}\)

\(\Rightarrow3S=4S-S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{29}}-\dfrac{1}{4}-\dfrac{1}{4^2}-...-\dfrac{1}{4^{30}}=1-\dfrac{1}{4^{30}}\)

\(\Rightarrow S=\dfrac{1-\dfrac{1}{4^{30}}}{3}\)

=>2A=4/2*6+4/6*10+4/10*14+...+4/42*46

=>2A=1/2-1/6+1/6-1/10+...+1/42-1/46=1/2-1/46=22/46

=>A=11/46

\(A=\dfrac{1}{6}+\dfrac{1}{30}+\dfrac{1}{70}+\dfrac{1}{126}+...+\dfrac{1}{966}\)

\(\Rightarrow\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{6}+\dfrac{1}{30}+\dfrac{1}{70}+...+\dfrac{1}{966}\right)\)

\(\Rightarrow2A=\dfrac{4}{2\times6}+\dfrac{4}{6\times10}+\dfrac{4}{10\times14}+\dfrac{4}{14\times18}+...+\dfrac{4}{42\times46}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{18}+...+\dfrac{1}{42}-\dfrac{1}{46}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{46}\)

\(\Rightarrow2A=\dfrac{11}{23}\)

\(\Rightarrow A=\dfrac{11}{23}:2\)

\(\Rightarrow A=\dfrac{11}{46}\)

a: 4/9+3/7=28/63+27/63=55/63

3/4+7/24=18/24+7/24=25/24

1/3+2/9+4/27=9/27+6/27+4/27=19/27

b: 5/6-3/8=20/24-9/24=11/24

7/15-11/30=14/30-11/30=3/30=1/10

2/3+1/6-7/12

=8/12+2/12-7/12

=3/12=1/4

c: 18/25*15/6=15/25*18/6=3*3/5=9/5

30/49:6/7=30/49*7/6=210/294=5/7

1/2*3/4:6/5=3/8*5/6=15/48=5/16

d: 8*3/5:12/5=24/5*5/12=2

4:9/5:10/3=4*5/9*3/10=2/3

Ta có một số phân tích sau :  \(a^4\)\(+\)\(4\)\(=\)\(\left(a^2-2a+2\right)\)\(\left(a^2+2a+2\right)\)

Nhân mỗi biểu thức trong ngoặc ở cả tử thức với  \(16\)\(=\)\(2^4\), ta được :

\(A\)\(=\)\(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)

\(A\)\(=\)\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)...\left(58^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)\left(12^4+4\right)...\left(60^4+4\right)}\)

Kết hợp với phân tích nêu trên, khi đó :

\(A\)\(=\)\(\frac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)\left(6^2-2.6+2\right)\left(6^2+2.6+2\right)....\left(58^2-2.58+2\right)\left(58^2+2.58+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)\left(8^2-2.8+2\right)\left(8^2+2.8+2\right)....\left(60^2-2.60+2\right)\left(60^2+2.60+2\right)}\)

\(\Rightarrow\)\(A\)\(=\)\(\frac{2.10.26.50.82.122....3250.3482}{10.26.50.82.122....3482.3722}\)\(=\)\(\frac{2}{3722}\)\(=\)\(\frac{1}{1861}\)

`1/4. 2/6. 3/8. 4/10........ 30/62. 31/64=2^x`

`=>\underbrace{1/2. 1/2. 1/2. 1/2..............1/2}_{\text{30 số 2}}. 1/64=2^x`

`=>(1/2)^{30}.(1/2)^{6}=2^x`

`=>(1/2)^{36}=2^x`

`=>2^{-36}=2^x`

`=>x=-36`

Vậy `x=-36`

các bạn thêm bằng 2^x hộ mk nha , mk quên

=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)

=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)

=>x=-36