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a,thay n=1 vào thì sẽ bằng 24 ko chia hết cho 10 nên đề sai
b, \(5^n\left(5^2+5^1+1\right)=5^n.31\)
\(\left(3^{n+2}-2^{n+2}+3^n-2^n\right)\)
\(=3^n.3^2-2^n.2^2+3^n-2^n\)
\(=\left(3^n.9+3^n\right)-\left(2^n.4+2^n\right)\)
\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n\left(9+1\right)-2^{n-1}.2\left(4+1\right)\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\left(ĐPCM\right)\)
Ta có :
3n+2−2n+2+3n−2n3n+2−2n+2+3n−2n =3n.32−2n.22+3n−2n3n.32−2n.22+3n−2n
=3n.9−2n.4+3n−2n3n.9−2n.4+3n−2n =3n.(9+1)−2n.(4+1)3n.(9+1)−2n.(4+1)
=3n.10−2n.5=3n.10−2n−1.2.53n.10−2n.5=3n.10−2n−1.2.5 = 3n.10−2n−1.103n.10−2n−1.10
=10.(3n−2n−1)⋮1010.(3n−2n−1)⋮10
⇒3n+2−2n+2+3n−2n⋮10⇒3n+2−2n+2+3n−2n⋮10 (ĐPCM)
TK NHA
3n + 2−2n + 2 + 3n−2n3n + 2−2n + 2 + 3n−2n =3n.32−2n.22 + 3n−2n3n.32−2n.22 + 3n−2n
=3n.9−2n.4 + 3n−2n3n.9−2n.4 + 3n−2n =3n.(9 + 1)−2n.(4 + 1)3n.(9 + 1)−2n.(4 + 1)
=3n.10−2n.5 = 3n.10−2n−1.2.53n.10−2n.5 = 3n.10−2n−1.2.5 = 3n.10−2n−1.103n.10−2n−1.10
=10.(3n−2n−1)⋮1010.(3n−2n−1)⋮10
⇒3n + 2−2n + 2 + 3n−2n⋮10⇒3n + 2−2n + 2 + 3n−2n⋮10 (ĐPCM)
TK NHA. chúc bn hok tốt @_@
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
\(3^{n+2}-2^{n+2}\)\(+3^n-2^n\)\(=3^n.3^2-2^n.2^2\)\(+3^n-2^n\)\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)\(=3^n.10-2^n.5\)\(=3^n.10-2^{n-1}.10\)\(=10\left(3^n-2^{n-1}\right)⋮10\)
\(\left(x-1\right)^2+\left(2x-y-3\right)^2\)\(+\left(y+z\right)^2=0\)
Có \(\left(x-1\right)^2\ge0\forall x;\)\(\left(2x-y-3\right)^2\ge0\forall x,y;\)\(\left(y+z\right)^2\ge0\forall y,z\)
Suy ra x-1=2x-y-3=y+z=0
=> \(\hept{\begin{cases}x=1\\2-y-3=0\\y+z=0\end{cases}}\)<=> \(\hept{\begin{cases}x=1\\y=-1\\z=1\end{cases}}\)
Bài 1:
Ta có: \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{99}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}\)
Vì \(A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\) nên \(A< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\)
\(3^{n+2}-2^{n+2}+3^n-2^n=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1}.10=10.\left(3^n-2^{n-1}\right)⋮10\)
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\times\left(3^2+1\right)-2^n\times\left(2^2+1\right)\)
\(=3^n\times10-2^n\times5\)
=> \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)