Question

[Phạm Thùy Dung]

1 : Chứng minh rằng : \(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\) chia hết cho 120 ( với \(x\inℕ\))

2 . Cho ba số a , b , c khác 0 thỏa mãn điều kiện :

\(\frac{3a+b+c}{a}+\frac{a+3b+c}{b}+\frac{a+b+3c}{c}\)

Tính giá trị biểu thức \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)

3 Tìm các cặp số nguyên \(\left(x;y\right)\)thỏa mãn \(x+2y=3xy+3\)

Answer 1

[Xyz OLM]

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

Answer 2

[Kudo Shinichi]

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 \(120⋮120\)

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

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