a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

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

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

a. 3x² - 2x - 1 = 0

<=> 3x² - 3x + x - 1 = 0

<=> 3x(x - 1) + (x - 1) = 0

<=> (3x + 1)(x - 1) = 0

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

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

b. \(\dfrac{x+1}{3}+\dfrac{2x+3}{5}=\dfrac{3}{4}\)

<=> \(\dfrac{20\left(x+1\right)}{60}+\dfrac{12\left(2x+3\right)}{60}=\dfrac{45}{60}\)

<=> 20x + 20 + 24x + 36 = 45

<=> 44x = -11

<=> x = \(-\dfrac{1}{4}\)

a) \(3x^2-2x-1=0\) \(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)

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

b) Pt\(\Rightarrow\)\(5\cdot4\left(x+1\right)+3\cdot4\cdot\left(2x+3\right)=3\cdot3\cdot5\)

       \(\Leftrightarrow44x=-11\Rightarrow x=-\dfrac{1}{4}\)

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)=3\)

\(\Leftrightarrow2x^2+x-x^3-2x^2+x^3-x+3=3\)

\(\Leftrightarrow3=3\)( Luôn đúng với mọi x )

Vậy phương trình nghiệm đúng với mọi x

b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)

\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)

\(\Leftrightarrow-3x^3+6x^2-3x-24=12x+12\)

\(\Leftrightarrow-3x^3+6x^2-3x-24-12x-12=0\)

\(\Leftrightarrow-3x^3+6x^2-15x-36=0\)

Đến đây xem lại đề bạn nhớ :D Tìm thì tìm được nhưng thấy nó sai sai kiểu gì í

c) \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)

\(\Leftrightarrow3x\left(x-2\right)+1\left(x-2\right)=2\left(-3x-5\right)-x\left(-3x-5\right)\)

\(\Leftrightarrow3x^2-6x+x-2=-6x-10+3x^2+5x\)

\(\Leftrightarrow3x^2-6x+x+6x-3x^2-5x=-10+2\)

\(\Leftrightarrow-4x=-8\)

\(\Leftrightarrow x=2\)

d) \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)

\(\Leftrightarrow x\left(x+5\right)+3\left(x+5\right)-x\left(x+7\right)=2x+8\)

\(\Leftrightarrow x^2+5x+3x+15-x^2-7x=2x+8\)

\(\Leftrightarrow x^2+5x+3x-x^2-7x-2x=8-15\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

a, \(x\left(2x-1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)=3\)

\(\Leftrightarrow2x^2-x-x^3-2x^2+x^3-x+3=3\)

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

b, \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)

\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)

\(\Leftrightarrow-3x-24+6x^2-3x^3=12x+12\)

\(\Leftrightarrow-15x-36+6x^2-3x^3=0\)

Lớp 8 chưa hc vô tỉ đâu ... vô nghiệm 

c, \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)

\(\Leftrightarrow3x^2-5x-2=-x-10+3x^2\)

\(\Leftrightarrow-4x+8=0\Leftrightarrow x=2\)

d, \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)

\(\Leftrightarrow x^2+8x+15-x^2-7x=2x+8\)

\(\Leftrightarrow x+15=2x+8\Leftrightarrow-x+7=0\Leftrightarrow x=7\)

a) \(2x\left(x-3\right)+3\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{2}\end{matrix}\right.\)

b) \(x\left(3x-1\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)

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

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

1.

\(y'=12x+\dfrac{4}{x^2}\)

2.

\(y'=\dfrac{3}{\left(-x+1\right)^2}\)

3.

\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)

4.

\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)

\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)

5.

\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)

6.

\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)

Bài 3:

a: =>3x^2-6x-x-3x^2=14

=>-7x=14

=>x=-2

b: \(\Leftrightarrow2x^2+10x-x-5-2x^2-9x-x-4.5=3.5\)

=>-x-9,5=3,5

=>-x=12

=>x=-12

c: =>\(3x-3x^2+9x=36\)

=>-3x^2+12x-36=0

=>x^2-6x+12=0(loại)

d: \(\Leftrightarrow3x^2-3x+x-1+4x-3x^2=5\)

=>2x=6

=>x=3