1. a, M = -\(\dfrac{1}{3}.\dfrac{141}{17}-\dfrac{39}{3}.\left(-\dfrac{1}{17}\right)\)

= -\(\dfrac{1}{17}.\dfrac{141}{3}-\dfrac{39}{3}.\left(-\dfrac{1}{17}\right)\)

= -\(\dfrac{1}{17}\left(\dfrac{141}{3}-\dfrac{39}{3}\right)\)

= -\(\dfrac{1}{17}.34\)

= -2

@Lê Thị Hồng Ngát

1. b, \(\dfrac{3}{4}+\dfrac{1}{4}x=7\)

<=> \(\dfrac{1}{4}x=\dfrac{25}{4}\)

<=> x = 25

@Lê Thị Hồng Ngát

1. b, N = -\(\dfrac{9}{16}.\dfrac{13}{3}-\left(-\dfrac{3}{4}\right)^2.\dfrac{19}{3}\)

= -\(\dfrac{9}{16}.\dfrac{13}{3}+\dfrac{9}{16}.\dfrac{19}{3}\)

= \(\dfrac{9}{16}\left(-\dfrac{13}{3}+\dfrac{19}{3}\right)\)

= \(\dfrac{9}{16}.2\)

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

@Lê Thị Hồng Ngát

\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{99.101}\right)\)

\(=\frac{2.2}{1.3}\frac{3.3}{2.4}.....\frac{100.100}{99.101}\)

\(=\frac{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}\)

\(=\frac{100.2}{101}=\frac{200}{101}\)

\(\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right)^2}< 0\)

\(\Rightarrow\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right).\left(x-2\right)}< 0\)

=> ( x - 3 ) . ( x - 5 ) và ( x - 2 ) . ( x - 2 ) trái dấu 

Mà ( x - 2 )² = ( x - 2 ) . ( x - 2 ) ≥ 0 ∀ x

 \(\Rightarrow\hept{\begin{cases}\left(x−3\right).\left(x+5\right)< 0\\\left(x−2\right).\left(x−2\right)>0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< −5;−5< x< 3\\x>2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< −5\\2< x< 3\end{cases}}\)

A = 13/21.2/11 + 13/21.9/11 + 8/21
= (13/21) + (13/21) + (8/21)
= (13 + 13 + 8)/21
= 34/21

B = (1 - 1/5)(1 - 2/5)(1 - 3/5)...(1 - 9/5)
= (4/5)(3/5)(2/5)(1/5)(0/5)(-1/5)(-2/5)(-3/5)(-4/5)
= 0

C = (1 - 1/2)(1 - 1/3)(1 - 1/4)...(1 - 1/50)
= (1/2)(2/3)(3/4)(4/5)...(49/50)
= 1/50

D = (2^2/1.3) * (3^2/2.4) * (4^2/3.5) * (5^2/4.6) * (6^2/5.7)
= (4/3) * (9/8) * (16/15) * (25/23) * (36/35)
= 0.979

2

\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\frac{25}{51}\)

\(S1=\frac{25}{102}\)

Nhầm ,chỉ có một + 1/3.5 thôi các bạn nhé

1)\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2\times\frac{502}{1005}\)

\(=\frac{1004}{1005}\)

tự làm tiếp nhé

1.= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

   = \(1-\frac{1}{101}\) = \(\frac{100}{101}\)

2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

   = \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

   = \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)

a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

b)\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+..+\frac{4}{2008.2010}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1004}-\frac{1}{1005}=1-\frac{1}{1005}=\frac{1004}{1005}\)

c)\(\frac{1}{2}.\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{9800}\right)=\frac{1}{4}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{400}\)