\(A=\dfrac{x^2+4x+3}{x+4}\cdot\dfrac{x^2+12x+11-9}{x+11}:\dfrac{x^2+2x-x-6}{x+2}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{x+4}\cdot\dfrac{x^2+12x+2}{x+11}\cdot\dfrac{x+2}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x^2+12x+2\right)\left(x+2\right)}{\left(x+4\right)\left(x+11\right)\left(x-2\right)}\)
Giải các pt sau:
a)\(x^2+\dfrac{4x^2}{\left(x+2\right)^2}=12\)
b) \(\dfrac{x^2}{3}+\dfrac{48}{x^2}=5\left(\dfrac{x}{3}+\dfrac{4}{x}\right)\)
c) \(\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}\)
d) \(\left(\dfrac{x-1}{x}\right)^2+\left(\dfrac{x-1}{x-2}\right)^2=\dfrac{40}{9}\)
e) \(x^2+\left(\dfrac{x}{x-1}\right)^2=8\)
g) \(x^3+\dfrac{1}{x^3}=6\left(x+\dfrac{1}{x}\right)\)
f) \(\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-12=0\)
Giải các pt sau:
a) \(x^2+\dfrac{4x^2}{\left(x+2\right)^2}=12\)
b) \(\dfrac{x^2}{3}+\dfrac{48}{x^2}=5.\left(\dfrac{x}{3}+\dfrac{4}{x}\right)\)
c) \(\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}\)
d) \(\left(\dfrac{x-1}{x}\right)^2+\left(\dfrac{x-1}{x-2}\right)^2=\dfrac{40}{9}\)
e) \(x^2+\left(\dfrac{x}{x+1}\right)^2=8\)
f) \(x^3+\dfrac{1}{x^3}=6\left(x+\dfrac{1}{x}\right)\)
giúp mình với ạ
Bài 1: giải các PT:
a, \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
b, \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
c, \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
d, \(\dfrac{x+\dfrac{2\left(3-x\right)}{5}}{14}-\dfrac{5x-4\left(x-1\right)}{24}=\dfrac{7x+2+\dfrac{9-3x}{5}}{12}+\dfrac{2}{3}\)
\(e,\dfrac{x-\dfrac{3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-2015}{2}+\dfrac{x-2014}{3}}{ }\)
Giải các phương trình sau:
a) \(x^2+\dfrac{4x^2}{\left(x+2\right)^2}=12\)
b) \(\dfrac{x^2}{3}+\dfrac{48}{x^2}=5\left(\dfrac{x}{3}+\dfrac{4}{x}\right)\)
c) \(\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}\)
d) \(\left(\dfrac{x-1}{x}\right)^2+\left(\dfrac{x-1}{x-2}\right)^2=\dfrac{40}{9}\)
e) \(x^2+\left(\dfrac{x}{x-1}\right)^2=8\)
f) \(\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-12=0\)
g) \(x^3+\dfrac{1}{x^3}=6\left(x+\dfrac{1}{x}\right)\)
Giải phương trình:
a,\(\dfrac{x+1}{x^2+x+1}\)-\(\dfrac{x-1}{x^2-x+1}\)=\(\dfrac{3}{x.\left(x^4+x^2+1\right)}\)
b,\(\dfrac{1}{4x^2-12x+9}\)-\(\dfrac{3}{9-4x^2}=\dfrac{4}{4x^2+12x+9}\)
c,\(\dfrac{2}{\left(1-3y\right).\left(3y+11\right)}=\dfrac{1}{9y^2-y+1}-\dfrac{3}{\left(3y+11\right)^2}\)
d,\(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
Giúp mk nha mn
Giải các phương trình :
a) \(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
b) \(\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)
d) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
1) \(\dfrac{7x-3}{x-1}\) = \(\dfrac{2}{3}\)
2) \(\dfrac{2\left(3-7x\right)}{1+x}\) = \(\dfrac{1}{2}\)
3) \(\dfrac{x^{2^{ }}-6}{x}\) = x + \(\dfrac{3}{2}\)
4) \(\dfrac{5}{3x+2}\) = 2x - 1
5) \(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}\) = 0
6) \(\dfrac{1}{x-2}\) + 3 = \(\dfrac{3-x}{x-2}\)
Giúp mk nha
Giải các phương trình
1/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
2/ \(\dfrac{1}{x^2-6x+8}+\dfrac{1}{x^2-10x+24}+\dfrac{1}{x^2-14x+48}=\dfrac{1}{9}\)
3/ \(\dfrac{1}{x^2-3x+3}+\dfrac{2}{x^2-3x+4}=\dfrac{6}{x^2-3x+5}\)
4/ \(\dfrac{6}{\left(x+1\right)\left(x+2\right)}+\dfrac{8}{\left(x-1\right)\left(x+4\right)}=1\)
5/ \(4\left(x^3+\dfrac{1}{x^3}\right)=13\left(x+\dfrac{1}{x}\right)\)
6/ \(\dfrac{4x}{4x^2-8x+7}+\dfrac{3x}{4x^2-10x+7}=1\)
7/ \(\dfrac{x^2-3x+5}{x^2-4x+5}-\dfrac{x^2-5x+5}{x^2-6x+5}=-\dfrac{1}{4}\)
8/ \(x.\dfrac{8-x}{x-1}\left(x-\dfrac{8-x}{x-1}\right)=15\)
Giải các phương trình sau :
a) \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2+1\right)}{x^2-4}\)
b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\)
c) \(\dfrac{2}{x-1}+\dfrac{2x+3}{x^2+x+1}=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
d) \(\dfrac{x^3-\left(x-1\right)^3}{\left(4x+3\right)\left(x-5\right)}=\dfrac{7x-1}{4x+3}-\dfrac{x}{x-5}\)
