Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow3^{2\left(x+1\right)}-5.3^{2x}=324\)
\(\Rightarrow3^2\left(3^{x+1}-5.3^x\right)=324\)
\(\Rightarrow9.3^x\left(3-5\right)=324\)
\(\Rightarrow3^x.\left(-2\right)=36\)
\(\Rightarrow3^x=-18=3^2.\left(-2\right)\)(vô lí vì 3x không chia hết cho 2)
c) \(\left(1-x\right)^5=32=2^5\)
\(\Rightarrow1-x=2\)
\(\Rightarrow x=-1\)
d) \(3.5^{2x+1}-3.25^x=300\)
\(\Rightarrow3\left(5^{2x}.5-5^{2x}\right)=300\)
\(\Rightarrow5^{2x}\left(5-1\right)=100\)
\(\Rightarrow5^{2x}.4=100\)
\(\Rightarrow5^{2x}=25=5^2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
a: \(=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3\)
b: \(M=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)
\(=3^{2\cdot1\cdot13\cdot12}=3^{312}\)
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
=> \(2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{16}{34}\)
=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
=> \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
=> \(1-\frac{1}{x+2}=\frac{16}{17}\)
=> \(\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
=> \(x+2=17\)
=> \(x=15\)
=>1/1-1/3+1/3-1/5+1/5-1/7+....+1/x-1/(x+2)=16/34
=>1/1-1/(x+2)=16/34
=>1/(x+2)=1-16/34
=>1/(x+2)=9/17
=>(x+2).9=17
=>(x+2)=17/9
=>x=17/9-2
=>x=-1/9(không là số tự nhiên)
vậy không có số tự nhiên x thoả mãn điều kiện bài toán
Ta có: \(5^{x-1}+3\cdot5^{x+1}=1900\)
\(\Leftrightarrow5^x:5+3\cdot5^x\cdot5=1900\)
\(\Leftrightarrow5^x\cdot\frac{1}{5}+15\cdot5^x=1900\)
\(\Leftrightarrow5^x\cdot\left(\frac{1}{5}+15\right)=1900\)
\(\Leftrightarrow5^x\cdot15,2=1900\)
\(\Leftrightarrow5^x=\frac{1900}{15,2}=125\)
\(\Leftrightarrow5^x=5^3\)
hay x=3
Vậy: x=3
a, Ta có \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
<=> \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)
<=> \(\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)-\left(\frac{x-3}{2009}-1\right)-\left(\frac{x-4}{2008}-1\right)=0\)
<=>\(\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> \(x-2012=0=>x=2012\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
=>\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}=2\cdot\frac{49}{99}\)
=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(1-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(\frac{2x}{2x+1}=\frac{98}{99}\)
=>2x = 98
=>x = 49
Lời giải:
$x(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7})< 1\frac{6}{7}$
$x(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7})< \frac{13}{7}$
$x(1-\frac{1}{7})< \frac{13}{7}$
$x.\frac{6}{7}< \frac{13}{7}$
$x< \frac{13}{7}: \frac{6}{7}=\frac{13}{6}$
Vì $x$ là số nguyên nên $x\leq 2$
Vậy $x$ là các số nguyên sao cho $x\leq 2$.
\(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\\\Rightarrow2^x\cdot2^2\cdot3^x\cdot3\cdot5^x=10800\\\Rightarrow(2^x\cdot3^x\cdot5^x)\cdot(2^2\cdot3)=10800\\\Rightarrow(2\cdot3\cdot5)^x\cdot(4\cdot3)=10800\\\Rightarrow30^x\cdot12=10800\\\Rightarrow30^x=10800:12\\\Rightarrow30^x=900\\\Rightarrow30^x=30^2\\\Rightarrow x=2\)
\(2^{x+2}.3^{x+1}.5^x=10800\\ \Leftrightarrow2^x.2^2.3^x.3.5^x=10800\\ \Leftrightarrow12.\left(2.3.5\right)^x=10800\\ \Leftrightarrow12.30^x=10800\\ \Leftrightarrow30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)