a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).

b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).

c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).

d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).

b, bạn xem lại đề 

c, đk : x khác 1 ; 3 

\(\Rightarrow x^2-8x+15+2x-2=x^2-4x+3\Leftrightarrow-2x=-10\Leftrightarrow x=5\left(tm\right)\)

d, đk: x khác -3 ; x khác 1 

\(\Rightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)

\(\Leftrightarrow2x^2+3x-5+x^2+2x-3=4+3x^2+8x-3\)

\(\Leftrightarrow-3x=5\Leftrightarrow x=-\dfrac{5}{3}\left(tm\right)\)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

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

\(\Leftrightarrow x=\dfrac{7}{2}\)

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

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a (x + 2) - x(x + 3) = 2

x + 2 - x(x + 3) - 2 = 0

x + x(x + 3) = 0

x(1 + x + 3) = 0

x(x + 4) = 0

x = 0 hoặc x + 4 = 0

*) x + 4 = 0

x = -4

Vậy x = -4; x = 0

b) (x + 2)(x - 2) - (x + 1)² = 7

x² - 4 - x² - 2x - 1 = 7

-2x - 5 = 7

-2x = 7 + 5

-2x = 12

x = 12 : (-2)

x = -6

c) 6x² - (2x + 1)(3x - 2) = 1

6x² - 6x² + 4x - 3x + 2 = 1

x + 2 = 1

x = 1 - 2

x = -1

d) (x + 2)(x + 3) - (x - 2)(x + 1) = 2

x² + 3x + 2x + 6 - x² - x + 2x + 2 = 2

6x + 8 = 2

6x = 2 - 8

6x = -6

x = -6 : 6

x = -1

e) 6(x - 1)(x + 1) - (2x - 1)(3x + 2) + 3 = 0

6x² - 6 - 6x² - 4x + 3x + 2 + 3 = 0

-x - 1 = 0

x = -1

`#3107.101107`

`1.`

`a,`

`(2x - 3)^2 = |3 - 2x|`

`=> (2x - 3)^2 = |2x - 3|`

`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

Vậy, `x \in {3/2; 2; 1}`

`b,`

`(x - 1)^2 + (2x - 1)^2 = 0`

`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

`c,`

`5 - x^2 = 1`

`=> x^2 = 4`

`=> x^2 = (+-2)^2`

`=> x = +-2`

Vậy, `x \in {-2; 2}`

`d,`

`x - 2\sqrt{x} = 0`

`=> x^2 - (2\sqrt{x})^2 = 0`

`=> x^2 - 4x = 0`

`=> x(x - 4) = 0`

`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy, `x \in {0; 4}`

`g,`

`(x - 1) + 1/7 = 0`

`=> x - 1 + 1/7 = 0`

`=> x - 6/7 = 0`

`=> x = 6/7`

Vậy, `x = 6/7.`

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1