Tìm x
/x-1/+/x+3/=5
CÁC BẠN GIÚP MIK VỚI MIK SẮP THI R
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
14 tháng 12 2018
\(\left(x-2013\right)^{2014}=1\)
\(\Rightarrow\orbr{\begin{cases}x-2013=1\\x-2013=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2013+1\\x=-1+2013\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2014\\x=2012\end{cases}}\)
Vậy x=2014 hoặc x=2012
hok tốt
31 tháng 7 2023
\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)
VN
Võ Ngọc Phương
CTVHS
VIP
5 tháng 9 2023
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
5 tháng 9 2023
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
9 tháng 10 2019
\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)
\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)
\(\rightarrow10x+80+15x+105=-6x\)
\(\Leftrightarrow31x+185=0\)
\(\Leftrightarrow x=-\frac{185}{31}\)
b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)
\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)
\(\rightarrow20x-160+15x-105=240+12-12x\)
\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)
TN
7
23 tháng 6 2020
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2}{5}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{10}\)
=> x+1=10
=>x=9
NK
9 tháng 3 2023
\(a,-\dfrac{12}{16}-\left(\dfrac{3}{4}-x\right)=-\dfrac{5}{3}\)
\(\dfrac{3}{4}-x=-\dfrac{12}{16}-\left(-\dfrac{5}{3}\right)\)
\(\dfrac{3}{4}-x=\dfrac{11}{12}\)
\(x=\dfrac{3}{4}-\dfrac{11}{12}\)
\(x=-\dfrac{1}{6}\)
\(b,x-\dfrac{3}{7}:\dfrac{9}{14}=-\dfrac{7}{3}\)
\(x-\dfrac{3}{7}=-\dfrac{7}{3}\times\dfrac{9}{14}\)
\(x-\dfrac{3}{7}=-\dfrac{3}{2}\)
\(x=-\dfrac{3}{2}+\dfrac{3}{7}\)
\(x=-\dfrac{15}{14}\)
\(c,-\dfrac{3}{4}x+\dfrac{5}{8}x=\dfrac{1}{3}\)
\(\left(-\dfrac{3}{4}+\dfrac{5}{8}\right)x=\dfrac{1}{3}\)
\(-\dfrac{1}{8}x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{1}{8}\right)\)
\(x=-\dfrac{8}{3}\)
NM
1
+) Nếu \(-3\le x\Leftrightarrow|x-1|=1-x\)
\(|x+3|=-x-3\)
\(pt\Leftrightarrow1-x-x-3=5\)
\(\Leftrightarrow-2x-2=5\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\frac{-7}{2}\left(tm\right)\)
+) Nếu \(-3< x< 1\Leftrightarrow|x-1|=1-x\)
\(|x+3|=x+3\)
\(pt\Leftrightarrow1-x+x+3=5\)
\(\Leftrightarrow4=5\) ( vô lí )
+) Nếu \(x\ge1\Leftrightarrow|x-1|=x-1\)
\(|x+3|=x+3\)
\(pt\Leftrightarrow x-1+x+3=5\)
\(\Leftrightarrow2x+2=5\)
\(\Leftrightarrow x=\frac{3}{2}\left(tm\right)\)
Vậy ....
Ta có:\(|x-1|\ge0\)
\(|x+3|\ge0\)
Theo bài:
\(|x-1|+|x+3|=5\)
\(\rightarrow x-1+x+3=5\)
\(\rightarrow\left(x+x\right)+[\left(-1\right)+3]=5\)
\(\rightarrow2x+2=5\)
\(\rightarrow2x=5-2\)
\(\rightarrow2x=3\)
\(\rightarrow x=3:2\)
\(\rightarrow x=\frac{3}{2}\)