\(\left(3k+1\right)^2=9k^2+6k+1chia3du1\)

\(\left(3k+2\right)^2=9k^2+12k+4chia3du1\)

\(\Rightarrow\left\{{}\begin{matrix}\left(16^2\right)^{1996}\equiv1\left(mod3\right)\\\left(17^2\right)^{1996}\equiv1\left(mod3\right)\\\left(13^2\right)^{1996}\equiv1\left(mod3\right)\end{matrix}\right.\Rightarrow\left(16^2\right)^{1996}+\left(17^2\right)^{1996}-\left(13^2\right)^{1996}+1\equiv1+1-1+1\equiv2\left(mod3\right)\Rightarrow dpcm\)

Ta co:

\(2001⋮3\Rightarrow2001^{2002}⋮3\) mà \(23\) chia 3 dư 2

\(\Rightarrow2001^{2002}+23\) chia 3 dư 2 \(\Rightarrow dpcm\)

b,

\(+,n=0\Rightarrow19^{2n}+5^n+2001=1+1+2001=2003\left(notscp\right)\)

\(+,n>0\Rightarrow19^{2n}+5^n+2001=361^n+5^n+2001=\left(...1\right)+\left(....5\right)+2001=\left(...7\right)\Rightarrow klscp\)

ĐK: a,b,c \(\ne\) 0

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Lại có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\)

Với \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}\)

\(\Rightarrow\) \(\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Rightarrow\) \(\dfrac{b+c}{bc}=0\) \(\Rightarrow\) b + c = 0 (vì bc \(\ne\) 0 do a,b,c \(\ne\) 0)

\(\Rightarrow\) b = -c \(\Rightarrow\) b⁵ = (-c)⁵ \(\Rightarrow\) b⁵ + c⁵ = 0

Thay b⁵ + c⁵ = 0 vào M ta được:

M = (a¹⁹ + b¹⁹).(b⁵ + c⁵).(c²⁰⁰¹ + a²⁰⁰¹)

M = (a¹⁹ + b¹⁹).0.(c²⁰⁰¹ + a²⁰⁰¹)

M = 0 (đpcm)

Chúc bn học tốt!

cảm ơn bn nha:))!!

phần a nhé

1/a+1/b+1/c=(a+b+c)(1/a+1/b+1/c)=3+(a/b+b/a)+(b/c+c/b)+(a/c+c/a)            do a+b+c=1

áp dụng bdt cosi cho các  so dương a/b,b/a,a/c,c/a,b/c,c/b

a/b+b/a >=2

b/c+c/b>=2

a/c+c/a>=2

cộng hết vào suy ra 1/a+1/b+1/c >=9       

https://hentaiz.net/

a) 2²⁰⁰² + 2²⁰⁰¹ = 2²⁰⁰¹(2 + 1) = 2²⁰⁰¹.3 = 2²⁰⁰⁰.(2.3) = 2²⁰⁰⁰.6 chia hết cho 6

b) 3¹⁰⁰⁰ + 3⁹⁹⁹  = 3⁹⁹⁹(3 + 1) = 3⁹⁹⁹.4 = 3⁹⁹⁸.(3.4) = 3⁹⁹⁸.12 chia hết cho 12

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)