Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B1: A=|x-13|+|x-2014|=|x-13|+|2014-x| \(\ge\) |x-13+2014-x| = 2001
Dấu "=" xảy ra khi \(\left(x-13\right)\left(2014-x\right)\ge0\Rightarrow13\le x\le2014\)
Vậy GTNN của A = 2001 khi 13\(\le\)x\(\le\)2014
B2
a, 3n+2-2n+2+3n-2n
=3n.32-2n.22+3n-2n
=3n(9+1)-2n(4+1)
=3n.10-2n.5
=3n.10-2n-1.10
=10(3n-2n-1) chia hết cho 10
b, \(\left(x-7\right)^{x+1}+\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}\Rightarrow\orbr{\begin{cases}x-7=0\\x-7=\pm1\end{cases}}\Rightarrow x\in\left\{6;7;8\right\}}\)
Bài 1 : \(3^{n+2}\)\(-2^{n+2}\)+ \(3^n-2^n\)= \(\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
= \(3^n\)\(\left(3^2+1\right)\) \(-2^n\left(2^2+1\right)\)= \(3^n\times10-2^{n-1}\times10\)
= 10 \(\times\left(3^n+2^{n+1}\right)\)
chia hết cho 10
Bài 2 :
\(A=75.\left(4^{2004}+4^{2003}+...+4^2+4+1\right)+25\) =\(75+25+75.4.\left(4^{2003}+4^{2003}+....+4^2+4\right)\)
= \(100+300.\left(4^{2003}+4^{2003}+...+4^2+4\right)\)
chia het cho 100
\(m-1⋮2m+1\)
\(\Rightarrow2m-2⋮2m+1\)
\(\Rightarrow2m+1-3⋮2m+1\)
\(\Rightarrow3⋮2m+1\)
tu lam
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot5\cdot2\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\)
Bài 1:
a) Sửa lại là: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\) nhé.
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
\(=3^n.\left(9+1\right)-2^{n-1}.2.\left(4+1\right)\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)\)
Vì \(10⋮10\) nên \(10.\left(3^n-2^{n-1}\right)⋮10.\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\left(đpcm\right)\left(\forall n\in N^X\right).\)
Chúc bạn học tốt!
2) \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=2^n.3^2-2^n.2^2+3^n-2^n\)
\(=2^n.9+2^n.4+3^n-2^n\)
\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
1) \(x+2y=3xy+3\)
\(\Rightarrow3xy+3-x-2y=0\)
\(\Rightarrow3xy-x+3-2y=0\)
\(\Rightarrow18xy-6x+18-12y=0\)
\(\Rightarrow6x\left(3y-1\right)+4-12y=-14\)
\(\Rightarrow6x\left(3y-1\right)-4\left(3y-1\right)=-14\)
\(\Rightarrow\left(6x-4\right)\left(3x-1\right)=-14\)
Bạn tự phân tích ra rồi tìm x, y nhé!
a, Ta có: 3xy - 5 = x2 + 2y
=> 3xy - x2 - 2y = 5
=> y.( 3x - 2 ) = 5 + x.x
=> y = \(\frac{5+x^2}{3x-2}\)
=> \(x^2+5⋮3x-2\)( vì y là số nguyên )
=> \(3x^2+15⋮3x-2\)
\(\Rightarrow x\left(3x-2\right)+15+2x⋮3x-2\)
\(\Rightarrow2x+15⋮3x+2\)
\(\Rightarrow6x+45⋮3x+2\)
\(\Rightarrow2.\left(3x+2\right)+41⋮3x+2\)
\(\Rightarrow41⋮3x+2\)
\(\Rightarrow3x+2\in\left\{-41;-1;1;41\right\}\)
\(\Rightarrow3x\in\left\{-43;-3;-1;39\right\}\)
VÌ 3x chia hết cho 3
\(\Rightarrow3x\in\left\{-3;39\right\}\)
\(\Rightarrow x\in\left\{-1;13\right\}\)
+) với x = -1 => y = -6/5 ( loại )
+) với x = 13 => y = 174/37 ( loại )
Vậy không tìm được ( x ; y ) thỏa mãn bài
b,
Xét \(3^{n+2}-2^{n+2}+3^n-2^n=3^n.9-2^n.4+3^n-2^n=3^n.\left(9+1\right)-2^n.\left(4+1\right)=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.2.5=3^n.10-2^{n-1}.10=10.\left(3^n-2^{n-1}\right)⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Vậy: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
1) Ta có: A=/x-2016/+/2017-x/nhỏ nhất thì 1 trong 2 GTTĐ phải bằng 0.
/x-2016/=0,nên x=2016;thay x=2016 vào /2017-x/:/2017-2016/=1.
Vậy GT nhỏ nhất là 1.