Tìm n \(\in\)N biết
3-3.3n+5.3n-3=486
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(5^{n+3}-5^{n+1}=5^{12}.120\Leftrightarrow5^{n+1}.\left(5^2-1\right)=5^{12}.5.24\)
\(\Leftrightarrow24.5^{n+1}=5^{13}.24\Leftrightarrow5^{n+1}=5^{13}\Leftrightarrow n+1=13\Leftrightarrow n=12\)
b) \(2^{n+1}+4.2^n=3.2^7\)
\(\Leftrightarrow2^n\left(2+4\right)=3.2^7\Leftrightarrow6.2^n=3.2^7\Leftrightarrow2^n=2^6\Leftrightarrow n=6\)
c) \(3^{n+2}-3^{n+1}=486\)
\(\Leftrightarrow3^{n+1}.\left(3-1\right)=486\Leftrightarrow2.3^{n+1}=486\Leftrightarrow3^{n+1}=243\)
\(\Leftrightarrow3^n=243:3=81=3^3\Leftrightarrow n=3\)
d) \(3^{2n+3}-3^{2n+2}=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.\left(3-1\right)=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.2=2.3^{10}\Leftrightarrow3^{2n+2}=3^{10}\Leftrightarrow2n+2=10\Leftrightarrow2n=8\Leftrightarrow n=4\)
\(\dfrac{3}{7}+\dfrac{a}{b}+\dfrac{2}{3}=\dfrac{1}{2}\)
\(\dfrac{3}{7}+\dfrac{a}{b}=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{3}{7}+\dfrac{a}{b}=-\dfrac{1}{6}\)
\(\dfrac{a}{b}=-\dfrac{1}{6}-\dfrac{3}{7}\)
\(\dfrac{a}{b}=-\dfrac{25}{42}\)
_____________
\(\dfrac{a}{b}-\dfrac{4}{9}+\dfrac{1}{10}=\dfrac{1}{7}\)
\(\dfrac{a}{b}-\dfrac{4}{9}=\dfrac{1}{7}-\dfrac{1}{10}\)
\(\dfrac{a}{b}-\dfrac{4}{9}=\dfrac{3}{70}\)
\(\dfrac{a}{b}=\dfrac{3}{70}+\dfrac{4}{9}\)
\(\dfrac{a}{b}=\dfrac{307}{630}\)
\(3\left(x-2\right)+4\left(x-5\right)=23\)
\(\Rightarrow3x-6+4x-20-23=0\)
\(\Rightarrow7x-49=0\)
\(\Rightarrow x=7\)
3(x-2)+4(x-5)=23
<=>3x-6+4x-20=23
<=>7x-26=23
<=>7x=49
<=>x=7
Vậy x=7
`3^1 .3^n + 5.3^(n+1)=16`
`3^(n+1) + 5.3^(n+1) =16`
`3^(n+1) . (1+5) =16`
`3^n . 3^1 = 8/3`
`3^n = 8/9`
`->` Không có n thỏa mãn.
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
Ta có: \(3^{n-3}+5.3^{n-3}=486\)
\(\Rightarrow3^{n-3}\left(1+5\right)=486\)
\(\Rightarrow3^{n-3}.6=486\)
\(\Rightarrow3^{n-3}=486:6\)
\(\Rightarrow3^{n-3}=81\)
\(\Rightarrow3^{n-3}=3^4\)
\(\Rightarrow3^n=3^{4+3}\)
\(\Rightarrow3^n=3^7\)
\(\Rightarrow n=7\)
tíck mình nha