5^6+5^7+5^8
=5^6.(1+5+5^2)
=5^6.31 chia hết cho 31
7^6+7^5-7^4
=7^4.(7^2+7-1)
=7^4.55 chia hết cho 11
BÀI 2:
a) \(5^6+5^7+5^8=5^6\left(1+5+5^2\right)=5^6.31\) \(⋮\)\(31\)
b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55\)\(⋮\)\(11\)
c) \(2^3+2^4+2^5=2^3.\left(1+2+2^2\right)=2^3.7\)\(⋮\)\(7\)
d) mk chỉnh đề
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+...+2^{58}\right)\)
\(=3\left(1+2^2+...+2^{58}\right)\)\(⋮\)\(3\)
a)Ta có : 56 + 57 + 58 = 55 .(5 + 52 + 53)
= 55 . 155 \(⋮\) 31 ( 155 \(⋮\)31 => 55 . 155 \(⋮\)31)
b) Ta có : 76 + 75 - 74 = 73 . (73 + 72 - 7)
= 73 . 385 \(⋮\)11 (385 \(⋮\) 11 => 73 . 385 \(⋮\)11)
Bài 3:
a) \(5^x+5^x.7=8.125\)
<=> \(5^x.\left(1+7\right)=8.125\)
<=> \(5^x.8=8.125\)
<=> \(5^x=125=5^3\)
<=> \(x=3\)
Vậy...
b) \(3^{x+2}-3^{x-1}=26.27\)
<=> \(3^{x-1}.3^3-3^{x-1}=26.27\)
<=> \(3^{x-1}.\left(3^3-1\right)=26.27\)
<=> \(3^{x-1}.26=26.27\)
<=> \(3^{x-1}=27=3^3\)
<=> \(x-1=3\)
<=> \(x=4\)
Vậy...
c) \(\left(3x-1\right)^2=121\)
<=> \(\orbr{\begin{cases}3x-1=11\\3x-1=-11\end{cases}}\)
<=> \(\orbr{\begin{cases}x=4\\x=-\frac{10}{3}\end{cases}}\)
Vậy...
Bài 4:
\(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}< 0\)
Nhận thấy: \(\left(2x-5\right)^{2000}\ge0;\) \(\left(3y+4\right)^{2002}\ge0\)
=> \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\) trái với đề bài
Vậy không tìm được x,y thỏa mãn bài ra
