NM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
22 tháng 5 2022
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
M
1 tháng 12 2019
Bài giải
b, \(x-5+\left|x-3\right|=4\)
\(\left|x-3\right|=4-x+5\)
\(\Rightarrow\orbr{\begin{cases}x-3=-4+x-5\\x-3=4-x+5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-x=-4-5+3\\x+x=4+5+3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\ne-6\text{ ( loại ) }\\2x=12\end{cases}}\)\(\Rightarrow\text{ }x=6\)
c, \(\sqrt{\left(x+7\right)^2}+\left(x^2-49\right)^{2012}=0\)
\(\left(x+7\right)+\left(x^2-49\right)^{2012}=0\)
\(\Rightarrow\hept{\begin{cases}x+7=0\\\left(x^2-49\right)^{2012}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-7\\x^2-49=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-7\\x^2=49\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-7\\x=\pm7\end{cases}}\)
\(\)\(\Rightarrow\text{ }x=-7\)
d, \(2\left|3-x\right|^{2017}+\left(y-x+1\right)^{2016}\le0\)
\(\text{Vì }\hept{\begin{cases}2\left|3-x\right|^{2017}\ge0\\\left(y-x+1\right)^{2016}\ge0\end{cases}}\) \(\Rightarrow\text{ Chỉ xảy ra trường hợp }2\left|3-x\right|^{2017}+\left(y-x+1\right)^{2016}=0\)
\(\Rightarrow\hept{\begin{cases}2\left|3-x\right|^{2017}=0\\\left(y-x+1\right)^{2016}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left|3-x\right|^{2017}=0\\y-x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3-x=0\\y-x+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y-3+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\y-2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)
WH
13 tháng 2 2018
a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
12 tháng 6 2017
a) \(\left|x+\frac{1}{5}\right|-4=-2\)
=) \(\left|x+\frac{1}{5}\right|=-2+4=2\)
=) \(x+\frac{1}{5}=2\)hoặc \(x+\frac{1}{5}=-2\)
=) \(x=2-\frac{1}{5}=\frac{9}{5}\); =) \(x=\left(-2\right)-\frac{1}{5}=\frac{-11}{5}\)
Vậy \(x=\left\{\frac{9}{5},\frac{-11}{5}\right\}\)
b)\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
=) \(2x-\frac{6}{5}x=\frac{-1}{2}+\frac{1}{5}\)
=) \(x.\left(2-\frac{6}{5}\right)=\frac{-3}{10}\)
=) \(x.\frac{4}{5}=\frac{-3}{10}\)
=) \(x=\frac{-3}{10}:\frac{4}{5}\)
=) \(x=\frac{-3}{8}\)
c) \(\left(x-3\right)^{x+2}-\left(x-3\right)^{x+8}=0\)
=) \(\left(x-3\right)^{x+2}.\left(1-6\right)=0\)
=) \(\left(x-3\right)^{x+2}=0:\left(1-6\right)=0\)
Mà chỉ có \(0^x=0\)
=) \(x-3=0\)
=) \(x=0+3\)
=) \(x=3\)
12 tháng 6 2017
a,
\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{5}\\x=-\frac{11}{5}\end{cases}}\)
b,
\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
\(\Rightarrow2x-\frac{6}{5}x=-\frac{1}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{4}{5}x=-\frac{3}{10}\Leftrightarrow x=-\frac{3}{8}\)
c,
\(\left[x-3\right]^{x+2}-\left[x-3\right]^{x+8}=0\)
=> [x-3]x + 2 = [x-3]x+8
=> x + 2 = x + 8
=> x không tồn tại
6 tháng 9 2020
Dài đấy :))
a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)
\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)
\(\Leftrightarrow\left|x-1\right|+8=9\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))
\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)
\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)
d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)
\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)
\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)
\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)
Vậy ta xét hai trường hợp sau :
1. \(x\ge-\frac{3}{16}\)
(*) <=>\(7x-2=4x+\frac{3}{4}\)
\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)
\(\Leftrightarrow3x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)
2. \(x< -\frac{3}{16}\)
(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)
\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)
\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)
\(\Leftrightarrow11x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)
Vậy x = 11/12
e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)
\(\Leftrightarrow x+1=4040\)
\(\Leftrightarrow x=4039\)
25 tháng 7 2022
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
`#3107`
`(x - 3)^5 = 4(x - 3)^3`
`=> (x - 3)^5 - 4(x - 3)^3 = 0`
`=> (x - 3)^3 * [ (x - 3)^2 - 4] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^3=0\\\left(x-3\right)^2-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^2=4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=\left(\pm2\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=2\\x-3=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=5\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 3; 5}.`
Thanks hen!