=>1/4x+300=800
=>1/4x=500
hay x=2000
\(\dfrac{1}{4}\left(\dfrac{1}{4}x+300\right)=200\)
\(\Leftrightarrow\dfrac{1}{4}x+300=800\)
\(\Leftrightarrow\dfrac{1}{4}x=500\)
\(\Leftrightarrow x=2000\)
1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
giải các phương trình sau
1, \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
2, \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
3, \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
Giải các pt sau:
1)\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+1}=\dfrac{3}{2-x}\)
2)\(\dfrac{3x+1}{1-3x}+\dfrac{3+x}{3-x}=2\)
3)\(\dfrac{8x-2}{3}=1+\dfrac{5-2x}{4}\)
4)
\(\dfrac{x}{x+1}-\dfrac{2x+3}{x}=\dfrac{-3}{x+1}-\dfrac{3}{x}\)
5)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
6)\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
giúp mình với cám ơn
a.\(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\) b.\(\dfrac{7}{x+2}=\dfrac{3}{x-5}\) c.\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) f.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
m.\(\dfrac{12}{1-9x^2}=\dfrac{1+3x}{1+3x}-\dfrac{1+3x}{1-3x}\) n.\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2}\) e.\(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)
d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) f.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) f.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
giải các phương trình sau
1, \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
2, \(\dfrac{3}{2+x}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
3, \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
