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\(3.3^{n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(\Leftrightarrow18.3^{2n+1}+3.3^n-2.3^{2n+3}+2.3^n=405\)
\(\Leftrightarrow54.3^{2n}+5.3^n-2.3^3.3^{2n}=405\)
\(\Leftrightarrow3^n=81\)
\(\Leftrightarrow n=4\)
ĐK \(n\ge0\)
Ta có \(3.3^{n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(\Leftrightarrow3^n\left(6.9.3^n+3\right)-2.3^n\left(27.3^n-1\right)=405\)
\(\Leftrightarrow54.3^{2n}+3.3^n-54.3^{2n}+2.3^n=405\Leftrightarrow5.3^n=405\)
\(\Leftrightarrow3^n=81=3^4\Leftrightarrow n=4\left(tm\right)\)
Vậy \(n=4\)
\(3.3^{n-1}.\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(\Rightarrow3.3^{n-1}.6.3^{n+2}+3.3.3^{n-1}-2.3^n.3^{n+3}+1.2.3^n=405\)
\(\Rightarrow3^{1+n-1}.6.3^n.3^2+3^{1+1+n-1}-2.3^n.3^n.3^3+3^n.2=405\)
\(\Rightarrow3^n.\left(6.3^2\right).3^n+3^{n+1}-\left(2.3^3\right).3^{n+n}+3^n.2=405\)
\(\Rightarrow\left(3^n.3^n\right).54+3^{n+1}-54.3^{2n}+3^n.2=405\)
\(\Rightarrow3^{2n}.54+3^{n+1}-3^{2n}.54+3^n.2=405\Rightarrow3^{n+1}+3^n.2=405\)
\(\Rightarrow3^n.3+3^n.2=405\Rightarrow3^n.5=405\Rightarrow3^n=81=3^4\Rightarrow n=4\)
\(\left(3^{n+1}-2.2^n\right)\left(3.3^n+2^{n+1}\right).3^{2n+2}+\left(8.2^{n-2}.3^{n+1}\right)^2\)
\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}+2^{n+1}\right).3^{2n+2}+\left(2^{n+1}.3^{n+1}\right)^2\)
\(=\left(3^{2n+2}-2^{2n+2}\right).3^{2n+2}+2^{2n+2}.3^{2n+2}\)
\(=3^{2\left(2n+2\right)}-2^{2n+2}.3^{2n+2}+2^{2n+2}.3^{2n+2}\)
\(=3^{2\left(2n+2\right)}=\left(3^{2n+2}\right)^2\).
Ta thấy \(\left(3^{2n+2}\right)^2\)luôn là 1 số chính phương với mọi n\(\in\)N
Nên ta có ĐPCM.
Lời giải:
Đặt biểu thức đã cho là $A$
Ta viết lại biểu thức thành:
\(A=(3^{n+1}-2^{n+1})(3^{n+1}+2^{n+1}).3^{2(n+1)}+(2^{n+1}.3^{n+1})^2\)
Đặt \(3^{n+1}=a; 2^{n+1}=b\Rightarrow A=(a-b)(a+b)a^{2}+(ba)^2\)
\(=(a^2-b^2)a^2+a^2b^2=a^4=(a^2)^2\)
Do đó biểu thức đã cho là một số chính phương.
Ta có đpcm.
vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
a) \(3\left(5-4n\right)+\left(27+2n\right)>0\)
\(\Leftrightarrow15-12n+27+2n>0\)
\(\Leftrightarrow42-10n>0\)
\(\Leftrightarrow-10n>-42\Leftrightarrow n< 4,2\)
Vậy \(S=\left\{n|n< 4,2\right\}\)
b) \(\left(n+2\right)^2-\left(n-3\right)\left(n+3\right)\le40\)
\(\Leftrightarrow n^2+4n+4-n^2+9\le40\)
\(\Leftrightarrow4n+13\le40\)
\(\Leftrightarrow4n\le27\Leftrightarrow n\le6,75\)
Vậy \(S=\left\{n|n\le6,75\right\}\)
\(3.3^{n-1}\left(6.3^{n+2}+3\right)-2.3^n\left(3^{n+3}-1\right)=405\)
\(\Leftrightarrow18.3^{2n+1}+3.3^n-2.3^{2n+3}+2.3^n=405\)
\(\Leftrightarrow54.3^{2n}+5.3^n-2.3^3.3^{2n}=405\)
\(\Leftrightarrow3^n=81\)
\(\Leftrightarrow n=4\)