Cho 5/(√x-2) ∈ Z. Tim x∈N thoa man.
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tim cac so m,n,p thoa man : m+n+p+8=2canm-1 + 4cann-2 +6canp-3
tim cac so x,y,z thoa man :canx+cany-1 +canz-2 = 1/2(x+y+z)
tim cac so x,y,z thoa man :x+y+z+4=2canx-2 +4cany-3+6canz-5
tim x;y;z thoa man :5x-3y/4=2z-3x/5=9y-10z/7;2y-z=132
tim x;y;z thoa man :x+3y/19=3y+9z/114=5z+15x;x+y+2z=-31
tim x thuoc Z thoa man : (x^2 - 20)(x^2-15)(x^2 - 10)(x^2 - 5 )<0
https://www.youtube.com/channel/UCjP80p-OtLhNnRs-R4Q7yjw
tim x thuoc Z thoa man : (x+3)(x-5)<0
cho các so tu nhien x,y,z,t nho nhat thoa man x/y=5/14;y/z=21/28;z/t=6/11.tim x,y,z,t
tim x,y,z,t thoa man x^2+y^2+z^2+t^2=x(y+z+t)
Tim x∈Z thoa man 33n+1=9n+2
Ta có: \(9^{n+2}=\left(3^2\right)^{n+2}=3^{2n+4}\)
=> \(3^{3n+1}=3^{2n+4}\)
=> 3n + 1 = 2n + 4
3n - 2n = 4 - 1
n = 3
Vậy n = 3.
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Tim x∈Z thoa man 33n+1 = 9n+2
\(3^{3n+1}=9^{n+2}\)
\(\Rightarrow3^{3n+1}=3^{2\left(n+2\right)}\)
\(\Rightarrow3n+1=2n+4\)
\(\Rightarrow3n-2n=4-1\)
\(\Rightarrow n=3\)
Vậy n=3 thì \(3^{3n+1}=9^{n+2}\)
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cho các so tu nhien x,y,z,t nho nhat thoa man x/y=5/14;y/z=21/28;z/t=6/11.tim x,y,z,t