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
NN
0
TA
18 tháng 12 2020
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
NT
1
10 tháng 11 2018
\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Cn lại lm tương tự nha e!
=.= hok tốt!!
HD
2
28 tháng 1 2018
a) \(\left(x^2+1\right)\left(2x-4\right)>0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x^2+1\right)>0\\\left(2x-4\right)>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2>-1\\2x>4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x>\sqrt{-1}\\x>2\end{cases}}}\)
Vậy \(x>2\)
b)\(\left(5x-15\right)\left(x^2+1\right)< 0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(5x-15\right)< 0\\x^2+1< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x< 15\\x^2< -1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< \sqrt{-1}\end{cases}}}\)
Vậy \(x< \sqrt{-1}\)
25 tháng 10 2023
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
12 tháng 6 2016
Bài này giống tìm nghiệm quá :
1) \(5\left(x-7\right)=0\)
\(\left(x-7\right)=0\div5\)
\(\left(x-7\right)=0\)
\(x=0+7\)
\(x=7\)
2) \(25\left(x-4\right)=0\)
\(\left(x-4\right)=0\div25\)
\(\left(x-4\right)=0\)
\(x=0+4\)
\(x=4\)
3) \(34\left(2x-6\right)=0\)
\(\left(2x-6\right)=0\div34\)
\(\left(2x-6\right)=0\)
\(2x=0+6\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
4) \(2007\left(3x-12\right)=0\)
\(\left(3x-12\right)=0\div2007\)
\(\left(3x-12\right)=0\)
\(3x=0+12\)
\(3x=12\)
\(x=12\div3\)
\(x=4\)
5) \(47\left(5x-15\right)=0\)
\(\left(5x-15\right)=0\div47\)
\(\left(5x-15\right)=0\)
\(5x=0+15\)
\(5x=15\)
\(x=15\div5\)
\(x=3\)
6) \(13\left(4x-24\right)=0\)
\(\left(4x-24\right)=0\div13\)
\(\left(4x-24\right)=0\)
\(4x=0+24\)
\(4x=24\)
\(x=24\div4\)
\(x=6\)
ND
2
23 tháng 8 2020
\(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\)
=> \(\left(\frac{1}{2}+\frac{1}{5}\right)x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x+\frac{3}{5}=0\)
=> \(\frac{7}{10}x=-\frac{3}{5}\)
=> \(x=\left(-\frac{3}{5}\right):\frac{7}{10}=\left(-\frac{3}{5}\right)\cdot\frac{10}{7}=\left(-\frac{3}{1}\right)\cdot\frac{2}{7}=-\frac{6}{7}\)
b) \(\left|2\frac{1}{2}+x\right|-\left(-\frac{2}{3}\right)=3\)
=> \(\left|\frac{5}{2}+x\right|+\frac{2}{3}=3\)
=> \(\left|\frac{5}{2}+x\right|=\frac{7}{3}\)
=> \(\orbr{\begin{cases}\frac{5}{2}+x=\frac{7}{3}\\\frac{5}{2}+x=-\frac{7}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{29}{6}\end{cases}}\)
c) \(\left|x+\frac{4}{15}\right|-\left|-3.75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=-\frac{8}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)
A
23 tháng 8 2020
a, \(\frac{1}{2}x+\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow\frac{7}{10}x+\frac{3}{5}=0\Leftrightarrow x=-\frac{6}{7}\)
b, đề sai
c, \(\left|\frac{x+4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\frac{x+4}{15}-3,75=-2,15\Leftrightarrow\frac{x+4}{15}=\frac{8}{5}\Leftrightarrow x+4=24\Leftrightarrow x=28\)
Nguyễn văn võ làm đúng rùi