a) \(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)

Vì \(x^2+2\ge2>0\forall x\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)

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

\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)

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

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

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

Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)

a, \(\left(x-3\right)\left(x^2+x-20\right)\ge0\)

\(\Leftrightarrow\) \(\left(x-3\right)\left(x-4\right)\left(x+5\right)\ge0\)

+) \(x-3=0\Leftrightarrow x=3\); \(x-4=0\Leftrightarrow x=4\); \(x+5=0\Leftrightarrow x=-5\)

+) Lập trục xét dấu f(x) (Bạn tự kẻ trục nha)

\(\Rightarrow\) Bpt có tập nghiệm S = \(\left[-5;3\right]\cup\) [4; \(+\infty\))

b, \(\dfrac{x^2-4x-5}{2x+4}\ge0\)

\(\Leftrightarrow\) \(\dfrac{\left(x-5\right)\left(x+1\right)}{2x+4}\ge0\)

+) \(x-5=0\Leftrightarrow x=5\); \(x+1=0\Leftrightarrow x=-1\); \(2x+4=0\Leftrightarrow x=-2\)

+) Lập trục xét dấu f(x) 

\(\Rightarrow\) Bpt có tập nghiệm S = (-2; -1] \(\cup\) [5; \(+\infty\))

c, \(\dfrac{-1}{x^2-6x+8}\le1\)

\(\Leftrightarrow\) \(\dfrac{\left(x-3\right)^2}{\left(x-4\right)\left(x-2\right)}\ge0\)

+) \(x-3=0\Leftrightarrow x=3\); \(x-4=0\Leftrightarrow x=4\); \(x-2=0\Leftrightarrow x=2\)

+) Lập trục xét dấu f(x)

\(\Rightarrow\) Bpt có tập nghiệm S = (\(-\infty\); 2) \(\cup\) (4; \(+\infty\))

Chúc bn học tốt!

\(a,\left(x-1\right)\left(x+2\right)=0\)

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

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

\(b,\left(x-2\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(c,\left(x+3\right)\left(x-5\right)=0\)

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

\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)

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

\(e,\left(x-4\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

\(f,\left(2x-1\right)\left(3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

`a,(x-1)(x+2)=0`

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

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

`b,(x -2)(x -5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

`c,(x +3)(x -5)=0`

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

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

`d,(x + 1/2)(4x + 4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)

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

`e,(x -4)(5x -10)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

`f,(2x -1)(3x +6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)

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

`g,(2,3x -6,9)(0,1x -2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)

a.(x - 1)(x + 2)= 0

<=> x-1=0 hoặc x+2=0

<=> x=1 hoặc x=-2

b.(x -2)(x -5)=0

<=> x-2=0 hoặc x-5=0

<=> x=2 hoặc x=5

c.(x +3)(x -5)=0

<=> x+3=0 hoặc x-5=0

<=> x=-3 hoặc x=5

d.(x + 1/2)(4x + 4)=0

<=> x+1/2=0 hoặc 4x+4=0

<=> x=-1/2 hoặc x=-1

e.(x -4)(5x -10)=0

<=> x-4=0 hoặc 5x-10=0

<=> x=4 hoặc x=2

f.(2x -1)(3x +6)=0

<=> 2x-1=0 hoặc 3x+6=0

<=> x=1/2 hoặc x=-2

g.(2,3x -6,9)(0,1x -2)=0

<=> 2,3x-6,9=0 hoặc 0,1x-2=0

<=> x=3 hoặc x=20

a: =>2x>=4

hay x>=2

b: =>-2x<=3

hay x>=-3/2

c: =>2x<=6

hay x<=3

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow x-1=3x-2\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

 \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)

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

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

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

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

c: =>(x-3)(x²+3x+5)=0

=>x-3=0

hay x=3

d: =>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x-3)(x-4)=0

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)

d: \(\Leftrightarrow3x^2-6x-2x+4=0\)

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

=>x=2 hoặc x=2/3

e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)

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

hay \(x\in\left\{0;3;-1\right\}\)

f: \(\Leftrightarrow x^2-5x-2+x=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow\left(x-2\right)^2=6\)

hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)