pt vt lại:
\(\dfrac{x+4}{x^2-3x+2}+\dfrac{x+1}{x^2-4}+5=\dfrac{2x+5}{x^2-4x+5}\)
pt này đk?
pt vt lại:
\(\dfrac{x+4}{x^2-3x+2}+\dfrac{x+1}{x^2-4}+5=\dfrac{2x+5}{x^2-4x+5}\)
pt này đk?
1.Giải các phương trình sau:
A. 1+14/(x-4)^2=-9/x-4
B.1+8x/1+2x-2x/2x-1+12x^2-9/1-4x^2=0
C.1/2x-6-3x-5/x^2-4x+3=1/2
a) 5-4x=3x-9
b) (x-4)(3x+9)=0
c) (x/(x+4))+(12/(x-4))=(4x+48)/(x*x-16)
d) 4-2x=7-x
e) (x+4)(8-4x)=0
f) (x/(x+5))+(11/(x-5))=(x+55)/(x*x-25)
g) ((3x+2)/2) - ((3x+1)/6)= (5/3) +2x
h) 2x-(3-5x)=4(x+3)
i) 3x -6+x=9-x
k) 2t-3+5t= 4t+12
m.n giúp mik ạ ...
Giải các PT sau:
a,4x-3/x-5=29/3
b,2x-1/5-3x=2
c,4x-5/x-1=2+x/x-1
d,7/x+2=3/x-5
e,2x+5/2x-x/x+5=0
f,12x+1/11x-4+10x-4/9=20x+17/18
a.4 / x+1 = 3 /x-2
b.x+2 / 2x-4 - 4x / x^2-4=0
c. 1 / x-3 -5/x = 3x+7 / x(x-3)
\(a,\dfrac{1}{x^2+3x+2}-\dfrac{3}{x^2-x-2}=\dfrac{-1}{x^2-4}\)
\(b,\dfrac{2x-1}{x^2+4x-5}+\dfrac{x-2}{x^2-10x+9}=\dfrac{3x-12}{x^2-4x-45}\)
Bài 1:Giải Phương trình
d) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
e)\(\frac{6}{x^{2^{ }}-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
g) \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)
h) \(\frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
a,x^2+2x/(x+1)^2+3-x^2-2x/(x-1)^2+3=16/x^4+4x^2+16
b,x^2/x^2+2x+2+x^2/x^2-2x+2=5(x^2-5)/x^4+4+25/4
1) \(\dfrac{4x+7}{x-1}\) = \(\dfrac{12x+5}{3x+4}\)
2) \(\dfrac{x}{x-1}\) - \(\dfrac{2x}{x^{2^{ }}-1}\) = 0
3) \(\dfrac{1}{3-x}\) - \(\dfrac{14}{x^2-9}\) = 1
4) \(\dfrac{x+1}{x-1}\) - \(\dfrac{x-1}{x+1}\) = \(\dfrac{4}{x^2-1}\)
5) x + \(\dfrac{1}{x}\) = x2 + \(\dfrac{1}{x^2}\)
6) \(\dfrac{x-1}{x^2+4}\) = \(\dfrac{x-1}{x+1}\)