Ta có: \(\dfrac{10x+3}{12}=\dfrac{15-8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{4\left(15-8x\right)}{36}\)
\(\Leftrightarrow30x+9=60-32x\)
\(\Leftrightarrow30x+32x=60-9=51\)
\(\Leftrightarrow62x=51\)
\(\Leftrightarrow x=\dfrac{51}{62}\)
Vậy: \(x=\dfrac{51}{62}\)
23) \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)
24) \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
25) \(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)
Giải các PT sau:
1) \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)
2) \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{15}\)
Giúp mình với!!!
\(a,\dfrac{1}{x^2+3x+2}-\dfrac{3}{x^2-x-2}=\dfrac{-1}{x^2-4}\)
\(b,\dfrac{2x-1}{x^2+4x-5}+\dfrac{x-2}{x^2-10x+9}=\dfrac{3x-12}{x^2-4x-45}\)
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) \(\dfrac{1}{4z^{2}-12z+9}-\dfrac{3}{9-4z^{2}}=\dfrac{4}{4z^{2}+12z+9}\)
b) \(\dfrac{2}{(1-3u)(3u+11)}=\dfrac{1}{9u^{2}-6u+1}-\dfrac{3}{(3u+11)^{2}}\)
c) \(\dfrac{4}{2x^{3}+3x^{2}-8x-12}-\dfrac{1}{x^{2}-4}-\dfrac{4}{2x^{2}+7x+6}+\dfrac{1}{2x+3}=0\)
10x+3/12=1+6+8x/9
\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\)
giải phương trình
\(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}=\dfrac{1}{6}\)
giai pt (giup toi voi)
1.\(\dfrac{4x}{4x^2-8x+7}+\dfrac{3x}{4x^2-10x+7}=1\)
2.\(\dfrac{2x}{x^2+3x-10}\)+\(\dfrac{5x}{x^2+7x-10}=4\)
