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a) \(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)=\left(3^n.3^2+3^n\right)-\left(2^n.2^2+2^n\right)\)
\(=\left[3^n.\left(3^2+1\right)\right]-\left[2^n.\left(2^2+1\right)\right]=\left(3^n.10\right)-\left(2^{n-1}.2.5\right)=\left(3^n.10\right)-\left(2^{n-1}.10\right)\)
Do: 3n . 10 chia hết cho 10 và 2n - 1 . 10 chia hết cho 10
=> ( 3n . 10 ) - ( 2n - 1 . 10 ) chia hết cho 10 => 3n + 2 - 2n + 2 + 3n - 2n chia hết cho 10
1) = 3n(32+1) - 2n(22+1)
2)A=m.n.p
\(\frac{m^2}{\frac{2^2}{5^2}}=\frac{n^2}{\frac{3^2}{4^2}}=\frac{p^2}{\frac{1^2}{6^2}}=\frac{m^2+n^2+p^2}{\frac{2^2}{5^2}+\frac{3^2}{4^2}+\frac{1^2}{6^2}}\)
3) \(\frac{a^2}{\text{\text{c}^2}}=\frac{\text{c}^2}{b^2}=\frac{a^2+\text{c}^2}{b^2+\text{c}^2}\)\(\frac{a^2}{\text{c}^2}=\frac{\text{c}^2}{b^2}=\frac{a^2+\text{c}^2}{\text{c}^2+b^2}\)
mà ab=c2
suy ra đpcm
Câu 1
4 p/s cộng thêm 1,p/s cuối trừ 4 rồi nhóm vs nhau
d/s la x= - 329
Câu 2
NHân vs 7 thành 7S rồi rút gọn là đc
Câu 1 :
a) \(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\) \(\Rightarrow x+329=0\Rightarrow x=-329\)
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=2^n.3^2-2^n.2^2+3^n-2^n\)
\(=2^n.9+2^n.4+3^n-2^n\)
\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
1/ \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{10}\)
\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\) (vì a + b + c = 2017)
\(\Rightarrow\left(\frac{c}{a+b}+1\right)+\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)=201,7\)
\(\Rightarrow M=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3=201,7\)
\(\Rightarrow M=198,7\)
2/
a, 3n+2 - 2n+2 + 3n + 2n
= 3n.32 + 3n - 2n.22 + 2n
= 3n.10 - 2n.5
= 3n.10 - 2n-1.10
= 10(3n - 2n-1 ) ⋮ 10
Ta có : 3n + 2 - 2n + 2 + 3n - 2n
= (3n + 2 + 3n) - (2n + 2 + 2n)
= 3n(32 + 1) - 2n - 1(23 + 2)
= 3n.10 - 2n - 1.10
= 10.(3n - 2n - 1)
Mà 3n - 2n - 1 thuộc Z
Nên 10.(3n - 2n - 1) chia hết cho 10
Vậy 3n + 2 - 2n + 2 + 3n - 2n chia hết cho 10
b)
a=3n+1+3n-1=3n(3+1)-1=3n*4-1
Để a chia hết cho 7 thì aEB(7)={1;7;14;28;35;...}
=>{3n*4}E{2;8;15;29;36;...}
=>3nE{9;...} => nE{3;...}
b=2*3n+1-3n+1=3n*(6-1)+1=3n*5+1
Để b chia hết cho 7 thì bEB(7)={1;7;14;28;35;...}
=>{3N*5}E{0;6;13;27;34;...}
=>3NE{0;...}
=>NE{0;...}
=>đpcm(cj ko chắc cách cm này)
Ta có:\(3^{n+2}-2^{n+2}+3^n-2^n\)
=\(\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
=\(3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
=\(3^n.10-2^n.5\)
=\(3^n.10-2^{n-1}.2.5\)
=\(3^n.10-2^{n-1}.10\)
=\(\left(3^n-2^{n-1}\right).10⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Nhớ tick cho mình nha!