Ta có : x = 0 không phải là nghiệm của phương trình
Xét \(x\ne0\), pt trở thành :
\(\Leftrightarrow\dfrac{2}{2x-5+\dfrac{3}{x}}+\dfrac{13}{2x+1+\dfrac{3}{x}}=6\)(1)
Đặt \(t=2x+\dfrac{3}{x}-2\)
\(\left(1\right)\Leftrightarrow\dfrac{2}{t-3}+\dfrac{13}{t+3}=6\)
\(\Leftrightarrow6t^2-15t-21=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=\dfrac{7}{2}\end{matrix}\right.\)
Tới đây đơn giản r *>
Giải phương trình:
a/ \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(x+3\right)}+\dfrac{2}{x+2}\)
b/ \(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)
c/ \(\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{2}{6-2x}\)
d/ \(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\)
Giải pt: \(\dfrac{3}{5x-1}+\dfrac{2}{3-5x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\dfrac{5+96}{x^2-16}=\dfrac{2x—1}{x+4}-\dfrac{3x-1}{4-x}\)
Giải pt:
\(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
Giải phương trình:
a) \(\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\)
b) \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
Giải Pt \(\dfrac{14}{20-6x-2x^2}+\dfrac{x^2+4x}{x^2+5x}-\dfrac{x+3}{2-x}+3=0\)
Giải các pt sau:
a) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\)
b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
\(a.x+\dfrac{2x+\dfrac{x-1}{5}}{3}\)
\(b.\dfrac{3x-1-\dfrac{x-1}{2}}{3}-\dfrac{2x+\dfrac{1-2x}{3}}{2}=\dfrac{\dfrac{3x-1}{2}-6}{5}\) giải pt
giải pt
\(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
giải các bất pt sau:
a, 4x-10<0
b, 2x+x+12\(\ge0\)
c, x-5\(\ge3-x\)
d, 7-3x>9-x
đ, 2x-(3-5x)\(\le4\left(x+3\right)\)
e, 3x-6+x<9-x
f, 2t-3+5t\(\ge\)4t+12
g, 3y-2\(\le2y-3\)
h,3-4x+24+6x\(\ge x+27+3x\)
i, 5-(6-x)\(\le4\left(3-2x\right)\)
k, 5(2x-3)- 4(5x-7)\(\ge19-2\left(x+11\right)\)
l, \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)
m, \(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\)
n, \(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\)
