\(-\dfrac{5}{6}x< 20\\ \Leftrightarrow x>20:-\dfrac{5}{6}\\ \Leftrightarrow x>-24\)
\(-\dfrac{5}{6}x< 20\)
\(-\dfrac{6}{5}\times\left(-\dfrac{5}{6}\right)x>-\dfrac{6}{5}\times20\)
\(x>-6\times4\)
\(x>-24\)
\(-\dfrac{5}{6}x< 20\\ \Leftrightarrow x>20:-\dfrac{5}{6}\\ \Leftrightarrow x>-24\)
\(-\dfrac{5}{6}x< 20\)
\(-\dfrac{6}{5}\times\left(-\dfrac{5}{6}\right)x>-\dfrac{6}{5}\times20\)
\(x>-6\times4\)
\(x>-24\)
\(\left(\dfrac{7x+5}{x^2+3x+2}-\dfrac{2x-3}{x+2}-\dfrac{x-1}{x+1}\right):\dfrac{20-6\sqrt{x}}{3x+6}\)
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 pt sau
a)\(\dfrac{60}{x}=\dfrac{4}{3}+\dfrac{60-x}{x+4}\)
b)\(\dfrac{100}{x}-\dfrac{100}{x+20}=\dfrac{5}{6}\)
c)\(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
Helppppp
Giải phương trình:
a) \(\dfrac{3x-2}{x^2-12x+20}-\dfrac{4x+3}{x^2+6x-16}=\dfrac{7x+11}{x^2-2x-80}\)
b) \(\dfrac{2x-5}{x^2+5x-36}-\dfrac{x-6}{x^2+3x-28}=\dfrac{x+8}{x^2+16x+63}\)
Cho A=\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{20}\). Chứng minh A>\(\dfrac{27}{20}\)
(3,0 điểm) Giải các phương trình sau:
a) $x-5=7-x$;
b) $3x-15=2x(x-5)$;
c) \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\).
Gải phương trình Sau
\(\dfrac{x^2+4x+6}{\text{x}+2}+\dfrac{x^2+16x+72}{x+8}\)=\(\dfrac{x^2+8x+20}{\text{x}+4}+\dfrac{x^2+12x+42}{x+6}\)
\(\dfrac{x+9}{x+4}\)- \(\dfrac{5}{x-4}\)=\(\dfrac{-20}{x^2-16}\)
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
Giải các phương trình sau:
a) \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
b) \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)