a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

T làm biếng lắm; làm C thôi

\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)

Làm tương tự ta được A > 1/15

câu a

\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)

\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)

@Ngô Tấn Đạt

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)

\(\Leftrightarrow3x-9=90-5+10x\)

\(\Leftrightarrow3x-9=10x+85\)

\(\Leftrightarrow3x-10x=85+9\)

\(\Leftrightarrow-7x=94\)

hay \(x=-\dfrac{94}{7}\)

Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)

b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)

\(\Leftrightarrow6x-4-60=9-6x-42\)

\(\Leftrightarrow6x-64=-6x-33\)

\(\Leftrightarrow6x+6x=-33+64\)

\(\Leftrightarrow12x=31\)

hay \(x=\dfrac{31}{12}\)

Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)

c) Ta có: \(3\left(x-1\right)+3=5x\)

\(\Leftrightarrow3x-3+3=5x\)

\(\Leftrightarrow3x-5x=0\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: S={0}

d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)

\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)

\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)

mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)

nên x+101=0

hay x=-101

Vậy: S={-101}

a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)

Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt

b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)

Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt

c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt

d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)

Vậy x = -101 là nghiệm của pt

e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

e) Ta có: \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\)

\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1=0\)

\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\)

\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\)

mà \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}>0\)

nên 100-x=0

hay x=100

Vậy: S={100}

f) Ta có: \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

mà \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}>0\)

nên x-100=0

hay x=100

Vậy: S={100}

Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:

a) \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{1}{1.2}\\\dfrac{1}{6}=\dfrac{1}{2.3}\\\dfrac{1}{12}=\dfrac{1}{3.4}\\...\end{matrix}\right.\)

Vậy số thứ 100 của dãy là: \(\dfrac{1}{100.101}=\dfrac{1}{10100}\)

Tổng: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

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

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

b) \(\left\{{}\begin{matrix}\dfrac{1}{6}=\dfrac{1}{\left(5.0+1\right)\left(5.1+1\right)}\\\dfrac{1}{66}=\dfrac{1}{\left(5.1+1\right)\left(5.2+1\right)}\\\dfrac{1}{176}=\dfrac{1}{\left(5.2+1\right)\left(5.3+1\right)}\\...\end{matrix}\right.\)

Vậy số thứ 100 của dãy là: \(\dfrac{1}{\left(5.99+1\right)\left(5.100+1\right)}=\dfrac{1}{248496}\)

Tổng: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{496.501}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{496.501}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)

\(=\dfrac{1}{5}.\dfrac{500}{501}\)

\(=\dfrac{100}{501}\)

Bài 2: Tính:

a) \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)

\(A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

\(A=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

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

Bài 2 :

a, Xét tử số : Đặt B = \(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}\)

Số số hạng của tử số là : ( 99 - 1 ) : 2 + 1 = 50 ( số )

=> Tử số có 50 phân số

Ta có : \(B=\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+\left(\dfrac{1}{5}+\dfrac{1}{95}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)\)

\(=\left(\dfrac{99}{99}+\dfrac{1}{99}\right)+\left(\dfrac{97}{3.97}+\dfrac{3}{3.97}\right)+\left(\dfrac{95}{5.95}+\dfrac{5}{5.95}\right)+...+\left(\dfrac{51}{49.51}+\dfrac{49}{49.51}\right)\)

\(=\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}\)

Xét mẫu số : Đặt C = \(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}\)

\(=\left(\dfrac{1}{1.99}+\dfrac{1}{99.1}\right)+\left(\dfrac{1}{3.97}+\dfrac{1}{97.3}\right)+...+\left(\dfrac{1}{49.51}+\dfrac{1}{51.49}\right)\)

\(=2.\dfrac{1}{1.99}+2.\dfrac{1}{3.97}+...+2.\dfrac{1}{49.51}\)

\(=2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)\)

Thay B và C vào A ta có :

\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)

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

Vậy A = 50

b, Xét mẫu số : Đặt C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)

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

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

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

Đặt D = 1 + 1 + ... + 1

Số số hạng của tổng D là : ( 99 - 1 ) : 1 + 1 = 99 ( số hạng )

\(\Rightarrow D=1.99=99\)

Thay D = 99 ta có :

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

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

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

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

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

Thay vào đề bài , ta có :

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

Vậy \(B=\dfrac{1}{100}\)

câu thứ 2 =0 vì (63.1,-21.3,6)=0

MIK muốn hỏi câu đầu tiên

b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)

Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé

1/

a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)

⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)

⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)

b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993

2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993

2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993

2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993

2.(1 − 1/x+1) = 3984/1993

1 − 1/x + 1= 3984/1993 :2

1 − 1/x+1 = 1992/1993

1/x+1 = 1 − 1992/1993

1/x+1=1/1993

<=>x+1 = 1993

<=>x+1=1993

<=> x+1=1993

<=> x = 1993-1

<=> x = 1992

Bài 1:

c: \(\Leftrightarrow x\cdot\dfrac{1}{2013}+1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}=2\)

\(\Leftrightarrow x\cdot\dfrac{1}{2013}+2-\dfrac{1}{2013}=2\)

=>x*1/2013=1/2013

=>x=1

d: \(\Leftrightarrow2x+8+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}=10\)

=>2x+1/2-1/3+1/3-1/4+...+1/9-1/10=2

=>2x+2/5=2

=>x+1/5=1

=>x=4/5