Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
⇔\(\frac{2}{x+2}-\frac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{5}{x^2-2x+4}=0\)
⇔\(\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
⇔\(2x^2-4x+8-2x^2-16-5\left(x+2\right)=0\)
⇔\(-4x+8-16-5x-10=0\)
⇔-9x-18=0
hay x=-2
Vậy: x=-2
\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{\left(x-4\right)}{x\left(x+2\right)}=0\)
\(\frac{1}{x-2}\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
giải pt
1,\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
2,\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
3,\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
4,\(\frac{2x}{x-1}+\frac{4}{x^2+2x-3=}=\frac{2x-5}{x+3}\)
5,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
6,\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
7,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
Bµi 5: Gi¶i PT sau.
\(a,\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
b,\(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
\(c,\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
d) (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
e) x4 + 2x3 + 4x2 + 2x + 1 = 0
\(f,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
giải phương trình sau :
a) 5-(x-6) = 4(3-2x) b) 2x(x+2)2-8x2 = 2(x-2)(x2+4)
c) 7-(2x+4) = -(x+4) d) (x+1)(2x-3) = (2x-1)(x+5
f) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
e) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
Giải các phương trình sau
a) \(\frac{7x-3}{x-1}=\frac{2}{3}\)
b) \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
c) \(\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
d) \(\frac{8-x}{x-7}-8=\frac{1}{x-7}\)
e) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
f)\(\frac{1}{x-1}+\frac{2}{x+1}=\frac{x}{x^2-1}\)
g) \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
h)\(5+\frac{76}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
i) \(\frac{90}{x}-\frac{36}{x-6}=2\)
k) \(\frac{1}{x}+\frac{1}{x=10}=\frac{1}{12}\)
l) \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)
m) \(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
n) \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
o)\(\frac{x}{2x+6}-\frac{x}{2x+2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
p) \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)
q) \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\)
r) \(\frac{x-1}{x}=\frac{1}{x+1}=\frac{2x-1}{x^2+x}\)
giải pt: a. (x - 2)(x+1)(x+3) = (x+3)(x+1)(2x-5)
b. \(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\frac{x+2}{x-5}-\frac{5}{x}=\frac{8}{x^2-2x}\)
\(\frac{2x+1}{4}-\frac{x-5}{3}\le\frac{4x-1}{12}+2\)
3) \(\frac{1-x}{x+1}-\frac{3+2x}{x+1}=0\)
13) \(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)
14) \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{20}{\left(x+1\right)\left(2-x\right)}\)
16) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
17) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
18) \(\frac{x-1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)
19) \(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)
20) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
1,Giải Pt
a,\(\frac{3x-7}{2}+\frac{x+1}{3}=-16\)
b,\(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
c,\(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5\left(5-2x\right)}{6}\)
e,\(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
