Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
219 - 7 . ( x+1) = 100
7 . ( x + 1 ) = 219 - 100 = 119
x+ 1= 119:7= 17
x = 17-1=16
(3x - 6 ). 3100 = 3103 2018x-3 = 2018
3x - 6 = 3103 : 3100 2018x-3 = 20181
3x - 6 = 33 =27 Do x-3=1 cho nên suy ra x=4
3x = 27 + 6
3x = 33
x = 33:3
x = 11
a) \(\left(x+1\right)-\frac{x+1}{3}=\frac{5\left(x+1\right)-1}{6}\)
\(\Leftrightarrow6\left(x+1\right)-2\left(x+1\right)=5\left(x+1\right)-1\)
\(\Leftrightarrow6x+6-2x-2=5x+5-1\)
\(\Leftrightarrow6x-2x-5x=5-1-6+2\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
b) \(\left(1-x\right)^2+\left(x+2\right)^2=2x\left(x-3\right)-7\)
\(\Leftrightarrow1-2x+x^2+x^2+4x+4=2x^2-6x-7\)
\(\Leftrightarrow2x^2+2x+5=2x^2-6x-7\)
\(\Leftrightarrow2x+6x=-7-5\)
\(\Leftrightarrow8x=-12\)
\(\Leftrightarrow x=-\frac{3}{2}\)
c) \(2+\frac{x-2}{2}-\frac{2x-4}{3}-\frac{5}{6}\left(2-x\right)=0\)
\(\Leftrightarrow2+\frac{x}{2}-1-\frac{2}{3}x+\frac{4}{3}-\frac{5}{3}+\frac{5}{6}x=0\)
\(\Leftrightarrow\frac{x}{2}-\frac{2}{3}x+\frac{5}{6}x=-2+1-\frac{4}{3}+\frac{5}{3}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{2}{3}\)
\(\Leftrightarrow x=-1\)
\(\left(x+2015\right)^7=\left(x+2015\right)^5\)
\(\left(x+2015\right)^7-\left(x+2015\right)^5=0\)
\(\left(x+2015\right)^5.\left(x+2015\right)^2-\left(x+2015\right)^5=0\)
\(\left(x+2015\right)^5\left[\left(x+2015\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2015\right)^5=0\\\left(x+2015\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=-2016;-2014\end{cases}}}\)
Vậy x = { - 2016; - 2014; 2015 }
a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)
b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)
c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)
d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)
a.\(\Rightarrow\left(\frac{3}{5}+x\right):\frac{2}{7}=\frac{3}{35}-\frac{2}{7}\)
\(\Rightarrow\left(\frac{3}{5}+x\right):\frac{2}{7}=-\frac{1}{5}\)
\(\Rightarrow\frac{3}{5}+x=-\frac{1}{5}.\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=-\frac{2}{35}\)
\(\Rightarrow x=-\frac{2}{35}-\frac{3}{5}\)
Vậy \(x=-\frac{23}{35}\).
b. => 5x-1=0 hoặc 2x-1/3=0
=> 5x=1 hoặc 2x=1/3
=> x=1/5 hoặc x=1/6
c. \(\Rightarrow\frac{1}{7}:x=\frac{3}{14}-\frac{3}{7}\)
\(\Rightarrow\frac{1}{7}:x=-\frac{3}{14}\)
\(\Rightarrow x=\frac{1}{7}:\left(-\frac{3}{14}\right)\)
Vậy \(x=\frac{-2}{3}\).
Các bạn ơi giúp mk với các bạn ơi mk sắp phải đi học rồi giúp mk với
Vì \(\left|x+\frac{3}{4}\right|\ge0;\left|y-\frac{1}{5}\right|\ge0;\left|x+y+z\right|\ge0\) với mọi x; y , z
nên để \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
thì \(\left|x+\frac{3}{4}\right|=\left|y-\frac{1}{5}\right|=\left|x+y+z\right|=0\)
=> \(x+\frac{3}{4}=0;y-\frac{1}{5}=0;x+y+z=0\)
+) x + 3/4 = 0 => x = -3/4
+) y - 1/5 = 0 => y =1/5
+) x + y + z = 0 => z = - x - y = 3/4 - 1/5 = 11/20
\(5\left(x-7\right)+2018^0=26\)
\(5x-35+1=26\)
\(5x-35=25\)
\(5x=60\)
\(x=12\)
\(5\left(x-7\right)+2018^0=26\)
\(\Rightarrow\left(5x-35\right)+1=26\)
\(\Rightarrow5x-35=25\)
\(\Rightarrow5x=60\)
\(\Rightarrow x=12\)
Vậy x = 12