Bài 1b:

\(\frac31\) + \(\frac33\) + \(\frac36\) + \(\frac{3}{10}\) + ...+\(\frac{3}{x\left(x+1\right):2}\) = \(\frac{2015}{336}\)

3.(\(\frac11+\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{x\left(x+1):2\right.})\) = \(\frac{2015}{336}\)

3.2(\(\frac12+\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)})=\) \(\frac{2015}{336}\)

6.(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\cdots+\frac{1}{x.\left(x+1\right)})\) = \(\frac{2015}{336}\)

6.(\(\frac11-\frac12\) + \(\frac12\)-\(\frac14\) +...+ \(\frac{1}{x}\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

6.(\(\frac11\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{336}\) : 6

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = 1 - \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = \(\frac{1}{2016}\)

\(x+1\) = 2016

\(x\) = 2016 - 1

\(x\) = 2015

Bài 2:

A = \(\frac{6n+1}{4n+3}\) (n ∈ Z\(^{-}\))

A ∈ Z khi và chỉ khi:

(6n + 1) ⋮ (4n + 3)

(12n + 2) ⋮ (4n + 3)

[3(4n + 3) - 7] ⋮ (4n + 3)

7 ⋮ (4n + 3)

(4n + 3) ∈ Ư(7) = {-7; -1; 1; 7}

n ∈ {- 5/2; -1; - 1/2; 1}

Nếu n = - 1 thì A = (-6 + 1)/(-4 + 3) = 5 (loại)

Nếu n = 1 thì: A = (6 + 1).(4+3) = 1 (loại)

Không có giá trị nào thỏa mãn đề bài hay n ∈ ∅

1. \(\frac{x}{y}=\frac{7}{17}\)

3. Có 6 cặp

4. 0 có cặp nào hết

Câu 2 mình không biết nha. Thông cảm

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\frac{3}{2}\Leftrightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{3}{2}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{1}{2}\)

\(\Leftrightarrow\frac{a+b+1}{ab}=\frac{1}{2}\Leftrightarrow2\left(a+b+1\right)=ab\Leftrightarrow2a+2b+2-ab=0\)

\(\Leftrightarrow2a-ab-4+2b+6=0\Leftrightarrow a\left(2-b\right)-2\left(2-b\right)=-6\)

\(\Leftrightarrow\left(a-2\right)\left(2-b\right)=-6\)

Đến đây chắc dễ rồi

b) \(\left(1+a\right).\frac{1}{1+b^2}=\left(1+a\right)\left(1-\frac{b^2}{1+b^2}\right)\)

\(\ge\left(1+a\right)\left(1-\frac{b^2}{2b}\right)=1+a-\frac{ab+b}{2}\)

Thiết lập hai BĐT còn lại tương tự và cộng theo vế được:

\(VT\ge6-\frac{ab+bc+ca+3}{2}\ge6-\frac{\frac{\left(a+b+c\right)^2}{3}+3}{2}\)

\(=6-\frac{3+3}{2}=3^{\left(đpcm\right)}\)

Dấu "=" xảy ra khi a = b = c = 1

1) Ta có : Đặt M = 3^{x + 1} + 3^{x + 2} + ... + 3^{x + 100}

= 3^x(3 + 3² + ... + 3¹⁰⁰) 

= 3^x[(3 + 3² + 3³ + 3⁴) + (3⁵ + 3⁶ + 3⁷ + 3⁸) + ... + (3⁹⁷ 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰)]

= 3^x[(3 + 3² + 3³ + 3⁴) + 3⁴.(3 + 3² + 3³ + 3⁴) + ... + 3⁹⁶.(3 + 3² + 3³ + 3⁴)]

= 3^x(120 + 3⁴.120 + .... + 3⁹⁶.120)

= 3^x.120.(1 + 3⁴ + .... + 3⁹⁶)

=> \(M⋮120\)(ĐPCM)

2) Ta có \(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)

\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)

\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

Nếu a + b + c = 0

=> a + b = - c

b + c = -a

c + a = -b

Khi đó P = \(\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

Nếu a + b + c \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)

Khi đó P = \(\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)

Vậy nếu a + b + c = 0 thì P = -3

nếu a + b + c  \(\ne\)0 thì P = 6

Ta có : 

\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)

\(=\left(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}\right)+...\)\(+\left(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}\right)\)

\(=3^x\left(3+3^2+3^3+3^4\right)+...+3^{x+96}\left(3+3^2+3^3+3^4\right)\)

\(=3^x.120+3^{x+4}.120+...+3^{x+96}.120\)

\(=120.\left(3^x+3^{x+4}+...+3^{x+96}\right)\)

Vì \(120⋮120\)

\(\Rightarrow120.\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)

\(\Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(\forall x\inℕ\right)\left(đpcm\right)\)