\(\dfrac{3x^2-18}{2x^2+z}=0\)
\(\Leftrightarrow3x^2-18=0\)
\(\Leftrightarrow3x^2=18\)
\(\Leftrightarrow x^2=6\)
\(\Leftrightarrow x=\pm\sqrt{6}\)
Khi đó, \(2.6+z\ne0\)
\(\Leftrightarrow z\ne-12\)
Vậy \(\left\{{}\begin{matrix}x=\pm\sqrt{6}\\z\ne-12\end{matrix}\right.\) thì thoả mãn
Giair phương trình:
\(\dfrac{3x}{x-2}-\dfrac{x}{x+5}+\dfrac{2x^2}{7x-10-x^2}=0\)
Giai phương trình sau :
a) \(\dfrac{4x-17}{2x^2+1}\) = 0
b)\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x+2}\) = 0
c) \(\dfrac{4}{x-2}\)-x+2 = 0
5.c) \(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x-1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
6.b) \(\dfrac{4}{2x^3+3x^2-8x-12}-\dfrac{1}{x^2-4}-\dfrac{4}{2x^2+7x+6}+\dfrac{1}{2x+3}=0\)
\(a,\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
b,\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)
c,\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)
Giải các phương trình
a) \(\dfrac{2x-5}{x+5}=3\)
b) \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
c) \(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
d) \(\dfrac{5}{3x+2}=2x-1\)
a. \(\dfrac{3x-4}{x+1}=2\)
b. \(\dfrac{2x^2-3}{x}=2x+\dfrac{3}{4}\)
c. \(\dfrac{6}{3x+2}\)=\(2x+3\)
Giải các phương trình sau:
a) \(\dfrac{1}{4z^{2}-12z+9}-\dfrac{3}{9-4z^{2}}=\dfrac{4}{4z^{2}+12z+9}\)
b) \(\dfrac{2}{(1-3u)(3u+11)}=\dfrac{1}{9u^{2}-6u+1}-\dfrac{3}{(3u+11)^{2}}\)
c) \(\dfrac{4}{2x^{3}+3x^{2}-8x-12}-\dfrac{1}{x^{2}-4}-\dfrac{4}{2x^{2}+7x+6}+\dfrac{1}{2x+3}=0\)
a) \(\dfrac{4x-8}{2x+1}\)=0 b) 2-\(\dfrac{x}{x-2}\)=\(\dfrac{x+1}{x+2}\) c)\(\dfrac{x+2}{x-3}\)=\(\dfrac{x}{x+1}\) d) \(\dfrac{3}{x-2}\)=\(\dfrac{1}{x}\) e) \(\dfrac{x-2}{x}\)=\(\dfrac{x-3}{x+2}\) f) \(\dfrac{2x}{x+1}\)-\(\dfrac{1}{x-2}\)=\(\dfrac{2x^2-16}{\left(x+1\right)\left(x-2\right)}\) g) \(\dfrac{3x}{x-1}+\dfrac{2}{x+1}=\dfrac{3x^2}{\left(x-1\right)\left(x+1\right)}\) h) \(\dfrac{1-x}{4}=\dfrac{2x+1}{5}\) i)\(\dfrac{2-3x}{3}=\dfrac{x+4}{4}\) m) \(\dfrac{4x+3}{5}=\dfrac{3-4x}{3}\) n) \(\dfrac{7-3x}{4}-2=\dfrac{x+5}{3}\)
\(a.x+\dfrac{2x+\dfrac{x-1}{5}}{3}\)
\(b.\dfrac{3x-1-\dfrac{x-1}{2}}{3}-\dfrac{2x+\dfrac{1-2x}{3}}{2}=\dfrac{\dfrac{3x-1}{2}-6}{5}\) giải pt
