Giải PT sau:
a, 3x - 7 = 0
b, 8 - 5x = 0
c, 3x - 2 = 5x + 8
d, \(\dfrac{3x-2}{3}\) = \(\dfrac{1-x}{2}\)
e, ( 5x + 1)(x - 3) = 0
f, (x + 1)(2x - 3) = 0
g, 4x(x + 3) - 5(x + 3) = 0
h, 8(x - 6) - 2x(6 - x) = 0
i, \(\dfrac{2}{x-1}\) + \(\dfrac{1}{x}\) = \(\dfrac{2x+5}{x^2-x}\)
k, \(\dfrac{3}{x+2}\) - \(\dfrac{2}{x-2}\) = \(\dfrac{2-x}{x^2-4}\)
m, \(\dfrac{3}{x}\) - \(\dfrac{2}{x-3}\) = \(\dfrac{4-x}{x^2-3}\)
n,\(\dfrac{3}{2x+10}\)+ \(\dfrac{2x}{x^2-25}\) = \(\dfrac{3}{x-5}\)
u, \(\dfrac{2}{x+3}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{x+4}{\left(x+3\right)\left(x-2\right)}\)
Tính:
\(a,\dfrac{x+3}{2x-1}-\dfrac{x^2-5}{4x^2-4x+1}-\dfrac{2x^3+5x^2-x-1}{8x^3-12x^2+6x-1}\)
\(b,\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
Giải các phương trình sau:
1. \(a,\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{2x-6}\)
\(b,\dfrac{1}{x-2}+\dfrac{5}{x+1}=\dfrac{3}{2-x}\)
\(c,\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
2. \(a,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(b,2x^2-6x+1\)
Giải phương trình sau:
\(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
\(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
Giải phương trình sau: \(1+\dfrac{x-2}{1-x}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
Giải các phương trình sau :
a,\(\dfrac{2}{2x+1}-\dfrac{3}{2x-1}=\dfrac{4}{4x^2-1}\)
b,\(\dfrac{2x}{x+1}+\dfrac{18}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)
c,\(\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
Giải bất phương trình sau:
a) 3.(4x + 1 ) -2x (5x+2) > 8x - 2
b) (4x2 -19x + 5 -2x - 3)2 > 0
c) x + 1 - \(\dfrac{x-1}{3}< \dfrac{2x+3}{2}+\dfrac{x}{3}+5\)
d) \(\dfrac{\left(3x-2\right)^2}{3}-\dfrac{\left(2x+1\right)^2}{3}< x\left(x+1\right)\)
e) 1- x .( 2x+1) = 0
g) \(\dfrac{4-x}{3x+5}\ge0\)
h) \(\dfrac{x+5}{x+7}-1>0\)
Mình cần gấp các bạn giúp mình với ạ. Mình cảm ơn nhiều lắm
Giải phương trình: \(A=\dfrac{1}{x^2-2x+2}+\dfrac{2}{x^2-2x+3}=\dfrac{6}{x^2-2x+4}\)
Giải phương trình :
a,\(\dfrac{2}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}=\dfrac{1}{x^2-3x+2}\)
b, \(\dfrac{2x+3}{x^2+3x+2}+\dfrac{6}{x^2-x-6}=\dfrac{2x-2}{x^2-2x-3}\)