\(\dfrac{x+2014}{2}+\dfrac{2\left(x+2014\right)}{7}=\dfrac{x+2014}{5}+\dfrac{x+2014}{6}\)
\(\left(x+2014\right)\left(\dfrac{1}{2}+\dfrac{2}{7}\right)=\left(x+2014\right)\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(\left(x+2014\right)\dfrac{11}{14}=\left(x+2014\right)\dfrac{11}{30}\)
Dấu ''=''↔x=-2014
bài 1 tìm x, y biết:
a,\(\dfrac{x}{2}=\dfrac{y}{5}\)và x+y=-14
b,\(\dfrac{x}{7}=\dfrac{y}{5}\)và x-y=8
Bài 2: Tìm x, y, z biết:
a,\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)và x+y+z=56
b,\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)và 2x+y-z=12
Tìm \(x\in Q\), biết rằng :
a) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
b) \(2x\left(x-\dfrac{1}{7}\right)=0\)
c) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
Tìm \(x\in Q\), biết rằng:
a) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
b) \(2x.\left(x-\dfrac{1}{7}\right)=0\)
c) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
bài 1: Tìm x biết:
a,\(x+\dfrac{1}{4}=\dfrac{3}{4}\)
b,\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)
c,\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)
d,\(2-x:\dfrac{3}{4}=\dfrac{1}{2}\)
e,\(\dfrac{11}{12}-\dfrac{2}{3}|x|=\dfrac{3}{8}\)
f,\(|2x-1|=5\)
lưu ý: Có câu có 2 trường hợp
Các bn giúp mk với
2x.(x-\(\dfrac{1}{7}\))=0
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
1.\(\dfrac{5}{7}+\dfrac{2}{3}\cdot x=\dfrac{3}{10}\)
2.\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
Tìm x :
\(7-\dfrac{1}{2}.\left(2x-5\right)-\dfrac{2}{3}\left(3x+1\right)=5\)
Bài 1: Tính
\(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)
Bài 2: Tìm x ∈ Q:
\(\left(x-\dfrac{4}{7}\right).\left(x+\dfrac{1}{2}\right)>0\)
Bài 1:
a) A=\(\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}\) + \(\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
b) Cho 3 số x, y, z là 3 số khác 0 thoả mãn điều kiện: \(\dfrac{y+z-x}{x}\) = \(\dfrac{z+x-y}{y}\) = \(\dfrac{x+y-z}{z}\)
Hãy tính giá trị biểu thức: B=(1+\(\dfrac{x}{y}\)) (1+\(\dfrac{y}{z}\)) (1+\(\dfrac{z}{x}\))
