\(\lim\limits_{x\rightarrow-\infty}\dfrac{x-5x^2+1}{x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}-5+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}=\dfrac{-5}{1}=-5\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}=\lim\limits_{x\rightarrow+\infty}\dfrac{5\left(\dfrac{2}{x^2}-1\right)^3\left(4+\dfrac{1}{x^2}\right)^2}{4+\dfrac{1}{x^{11}}-\dfrac{6}{x^{13}}}=\dfrac{5.\left(-1\right)^3.4^2}{4}=-20\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{4x-\sqrt{9x^2+x}}{3-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4-\sqrt{9+\dfrac{1}{x}}}{\dfrac{3}{x}-1}=\dfrac{4-3}{-1}=-1\)

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

a: Ta có: \(3x-\left(3x+2\right)=x+3\)

\(\Leftrightarrow x+3=-2\)

hay x=-5

b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)

\(\Leftrightarrow15x-3+8x-4=18x\)

\(\Leftrightarrow5x=7\)

hay \(x=\dfrac{7}{5}\)

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)

\(\Leftrightarrow3+x-5=6x-4\)

\(\Leftrightarrow x-2-6x+4=0\)

\(\Leftrightarrow-5x+2=0\)

\(\Leftrightarrow-5x=-2\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)

c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)

\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)

\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x+\dfrac{9}{4}=0\)

\(\Leftrightarrow x=-\dfrac{9}{4}\)

Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)

d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)

\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)

\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)

\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)

\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)

Vậy: \(S=\varnothing\)

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)

\(\Leftrightarrow2x-8+12x=4x-2\)

\(\Leftrightarrow10x=6\)

hay \(x=\dfrac{3}{5}\)

2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)

\(\Leftrightarrow15x-6-30=10-20x\)

\(\Leftrightarrow35x=46\)

hay \(x=\dfrac{46}{35}\)

3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)

\(\Leftrightarrow3x-6-4=6x-6\)

\(\Leftrightarrow-3x=4\)

hay \(x=-\dfrac{4}{3}\)

1)\(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)

\(\Leftrightarrow\dfrac{\left(x-4\right).2}{3.2}+\dfrac{2x.6}{6}=\dfrac{4x-2}{6}\)

\(\Rightarrow2x-8+12x=4x-2\\ \Leftrightarrow10x=6\\ \Leftrightarrow x=\dfrac{3}{5}\)

1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)

ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn