Ta có :
\(\dfrac{-1,64}{8,51}\)=\(\dfrac{\left(x\right)}{-3,11}\)
<=>/x/=\(\dfrac{-1,64\left(-3,11\right)}{8,51}\)=0,5993419506
<=>x=-0,5993419506 và 0,5993419506
theo mình là vậy
Ta có :
\(\dfrac{-1,64}{8,51}\)=\(\dfrac{\left(x\right)}{-3,11}\)
<=>/x/=\(\dfrac{-1,64\left(-3,11\right)}{8,51}\)=0,5993419506
<=>x=-0,5993419506 và 0,5993419506
theo mình là vậy
a,\(\left(3x-2\right):1\dfrac{2}{5}=2\dfrac{3}{7}:2\dfrac{3}{5}\)
b,\(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)
c,\(\dfrac{x+3}{8}=\dfrac{2}{x-3}\)
d,\(\dfrac{-1,64}{8,51}=\dfrac{\left|x\right|}{-3,11}\)
a) Tìm tập hợp các số nguyên x, biết rằng\(4\dfrac{5}{9}:2\dfrac{5}{18}-7< x< \left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\)
b) tìm x, biết \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+\left|x+\dfrac{1}{12}\right|+\left|x+\dfrac{1}{20}\right|+....+\left|x+\dfrac{1}{110}\right|-11x\)
c)Tính gt biểu thức \(C=2x^3-5y^3+2015\) tại x,y thỏa mãn \(\left|x-1\right|+\left(y+2\right)^{20}=0\)
Tìm x biết :
\(\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{4}\right|+....+\left|x-\dfrac{1}{10}\right|=2x\)
Tìm x biết: \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+\left|x+\dfrac{1}{12}\right|+\left|x+\dfrac{1}{20}\right|+...+\left|x+\dfrac{1}{110}\right|=11x\)
Tìm x.
\(1,\dfrac{3}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)=\dfrac{1}{4}\)
\(2,3\left(x-2\right)-4\left(x+2\right)=x+2\)
\(3,4x\left(x-1\right)+4x-2\left(x+1\right)=-2\)
\(4,x\left(x+2\right)-3\left(x-1\right)=3\left(x+1\right)\)
Tìm x biết
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+\left|x+\dfrac{1}{12}\right|+\left|x+\dfrac{1}{20}\right|+...+\left|x+\dfrac{1}{110}\right|=11x\)
Tìm x, y, zϵ R biết: \(\left(4x^2-4x+1\right)^{2022}+\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
Tìm x, y biết :
\(\left|x+3\right|+\left|x-1\right|=\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\)
Tìm x, biết \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+\left|x+\dfrac{1}{3\cdot4}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=100x\)