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3.32.33.34......3x=3190
=>3(1+2+3+...........+x)=3190
=>1+2+3+.........+x=190
=>\(\frac{x.\left(x+1\right)}{2}\)=180
=>x.(x+1)=190.2
=>x.(x+1)=380
=>x=19
![](https://rs.olm.vn/images/avt/0.png?1311)
3.3^2.3^3.............3^x=3^190
3^1+2+3+4+....+x=3^190
nên 1+2+3+.........+x=190
hay (x+1).x :2 =190 nen 190.2= (x+1) . x hay 380 =19.20
vay x=19
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=>31+2+3+4+...+x=3190
=>1+2+3+4+...+x=190
=>(x+1).x:2=190
=>(x+1).x=380
mà 380 chỉ có thể ptích thành: 380=20.19
=>x+1=20 và x=19
vậy x=19
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\cdot3^2\cdot3^3\cdot3^4\cdot....\cdot3^x=3^{190}\)
\(\Leftrightarrow3^{1+2+3+...+x}=3^{190}\)
\(\Leftrightarrow1+2+3+...+x=190\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2}=190\Leftrightarrow x\left(x+1\right)=380\)
\(\Leftrightarrow x^2+x-380=0\)
\(\Leftrightarrow x^2-19x+20x-380=0\)
\(\Leftrightarrow x\left(x-19\right)+20\left(x-19\right)=0\)
\(\Leftrightarrow\left(x-19\right)\left(x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-19=0\\x+20=0\end{matrix}\right.\)\(\Leftrightarrow x=19\left(x>0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(3^{1+2+3+..+x}=3^{3.12}\Leftrightarrow\frac{x\left(x+1\right)}{2}=36\)
\(\Leftrightarrow x.\left(x+1\right)=72=8.9\Leftrightarrow x=8\)
b. ta có
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2016}}=\left(\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)+1-\frac{1}{5^{2017}}\)
\(=A+1-\frac{1}{5^{2017}}\Rightarrow4A=1-\frac{1}{5^{2017}}< 1\Rightarrow A< \frac{1}{4}\)
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3S= 3+2.32+3.33+...+101.3101
<=> 2S= 101.3101-(3100+399+398+....+3)-1 (1)
Ta có
A=3100+399+...+3
<=> 3A=3101+3100+...+32
<=> A=\(\frac{3^{101^{ }}-3}{2}\)(2)
Thay (2) vào (1) ta có
S= \(\frac{101.3^{101}-\frac{3^{101}-3}{2}-1}{2}\)
<=> S=\(\frac{3^{101}.201-1}{2}.\frac{1}{2}\)=\(\frac{3^{101}.201-1}{4}\)
Mik nghĩ vậy k bt đúng k
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
eo ôi t làm rồi mà bị xoá :v thôi t hướng dẫn :v
Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)
=> 1+2+3+4+5+....+x = 190
x(x+1) = 190.2 = 380
x(x+1) = 19.(19 + 1)
VẬy x = 19