\(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-8x+7=x^2-7x+12\)
\(\Leftrightarrow-x=5\Leftrightarrow x=-5\)
Vậy x = -5
Câu1 : Tìm x biết :
a) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
b)\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
c)\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
BT3: Tìm x, biết
3) \(\dfrac{3}{4}.x-\dfrac{5}{3}.x=\dfrac{7}{12}\)
4) \(x-\dfrac{1}{7}=\) \(\dfrac{\dfrac{4}{2011}-\dfrac{4}{2013}+\dfrac{4}{99}}{\dfrac{5}{2011}-\dfrac{5}{2013}+\dfrac{5}{99}}+\dfrac{1}{2}\)
Các bạn giúp mk nhé
BT3: Tìm x, biết:
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
Bài 1: Tìm x,y thuộc Z biết:
\(a,\dfrac{x}{7}=\dfrac{9}{y}và\) x > y \(b,\dfrac{-2}{x}=\dfrac{y}{5}\) và x < 0 < y
Bài 2: Tìm x,y biết:
\(\dfrac{x-4}{y-3}=\dfrac{4}{3}\) và x - y = 5
BT: Tìm x, biết:
a) \(2\dfrac{2}{3}x+3\dfrac{3}{4}x=\dfrac{385}{24}\)
b) \(-2\dfrac{2}{3}:x=1\dfrac{7}{9}:0.8\)
c) \(140\%+\dfrac{5}{3}x+2x=2280\)
d) \(-3\dfrac{1}{3}\cdot\left|x\right|=1\dfrac{1}{2}-2\dfrac{2}{3}\)
Tìm x, biết:
1. \(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\)
2. x. ( x + 1) = 0
3. 2x - 16 = 40 + x
4. x : \(\dfrac{5}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{-2}{5}\)
5. 3\(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) x = \(\dfrac{2}{3}\)
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
7. \(\dfrac{1}{3}+\dfrac{1}{2}:x=\) -0,25
\(\dfrac{-3}{4}x\dfrac{4}{7}+\dfrac{-3}{5}x\dfrac{3}{7}-\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}x\dfrac{5}{6}-\dfrac{7}{12}\right)x\left(\dfrac{-12}{7}\right)\)
\(1,4x\dfrac{15}{19}x\left(\dfrac{4}{5}x\dfrac{-2}{3}\right)+\dfrac{7}{15}\)
Tính nha
a)x=\(\dfrac{1}{4}+\dfrac{2}{13}\)
b)\(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
c)\(\dfrac{-8}{3}+\dfrac{-1}{3}\le x\le\dfrac{-2}{7}+\dfrac{-5}{7}\)
d)\(\dfrac{-5}{6}+\dfrac{8}{3}\dfrac{29}{-x}\le x\le\dfrac{-1}{2}+2+\dfrac{5}{2}\)
tim x , biet :
b, |\(\dfrac{5}{4}\)x-\(\dfrac{7}{2}\)|=|\(\dfrac{5}{8}\)x+\(\dfrac{3}{5}\)|=0
c, |\(\dfrac{7}{5}\)x+\(\dfrac{2}{3}\)|=|\(\dfrac{4}{3}\)x-\(\dfrac{1}{4}\)|
d, |\(\dfrac{7}{8}\)x+\(\dfrac{5}{6}\)|-|\(\dfrac{1}{2}\)x+5|=0
