\(1,\\ a,ĐK:m\ne1\\ \Delta=49+48\left(m-1\right)=48m+1\\ \text{PT vô nghiệm }\Leftrightarrow48m+1< 0\Leftrightarrow m< -\dfrac{1}{48}\\ \text{PT có nghiệm kép }\Leftrightarrow48m+1=0\Leftrightarrow m=-\dfrac{1}{48}\\ \text{PT có 2 nghiệm phân biệt }\Leftrightarrow48m+1>0\Leftrightarrow m>-\dfrac{1}{48};m\ne1\)

\(b,\Delta=4\left(m-1\right)^2+4\left(2m+1\right)=4m^2+8>0,\forall m\\ \text{Vậy PT có 2 nghiệm phân biệt với mọi m}\\ 2,\\ \text{PT có 2 nghiệm phân biệt }\)

\(\Leftrightarrow\Delta=4\left(m+1\right)^2-4\left(m^2-1\right)>0\\ \Leftrightarrow4m^2+8m+4-4m^2+4>0\\ \Leftrightarrow8m+8>0\\ \Leftrightarrow m>-1\)

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)

\(\sqrt{9.\left(x-1\right)^2}-12=0\)

=> 3.(x - 1) - 12 = 0

=> 3x - 15 = 0

=> 3x = 15

=> x = 5

b) \(\sqrt{4.\left(3-x\right)}=16\) (ĐKXĐ: x ≤ 3)

\(\Rightarrow\sqrt{3-x}=8\)

=> 3 - x = 64

=> x = -61

a) Khi m = 0 thì phương trình trở thành:

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

\(\Leftrightarrow x^2+2\cdot-2x-0=0\)

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

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

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

b) Ta có: 

\(\left|x_1\right|-\left|x_2\right|=6\)

\(\Leftrightarrow x^2_1+x_2^2-2\left|x_1x_2\right|=36\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|=36\)

Mà: \(x_1+x_2=-2\left(m-2\right)=4-2m\)

\(x_1x_2=-m^2\)

\(\Leftrightarrow\left(4-2m\right)^2-2\cdot-m^2-2\cdot m^2=36\)

\(\Leftrightarrow16-16m+4m^2+2m^2-2m^2=36\)

\(\Leftrightarrow\left(4-2m\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4-2m=6\\4-2m=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2m=-2\\2m=10\end{matrix}\right.\)

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

\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)

\(b,\left(10-x\right)\left(3-x\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)

\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)

\(a,\left(x+12\right)\left(x+6\right)>0\) \(khi\) \(x>6\Rightarrow x\in\left\{7,8,9,...\right\}\)

\(b,\left(10-x\right)\left(3-x\right)< 0\) \(khi\) \(x< 10\Rightarrow x\in\left\{9,8,7,...\right\}\)

Lời giải:
a. $2x+7>0$

$\Leftrightarrow x> \frac{-7}{2}$

b. 

$-5x+12<17$

$\Leftrightarrow -5x< 5$

$\Leftrightarrow 5+5x>0$

$\Leftrightarrow 5x>-5$

$\Leftrightarrow x>-1$

c. 

$-3x+5>-5x+2$

$\Leftrightarrow (-3x+5)-(-5x+2)>0$

$\Leftrightarrow 2x+3>0$

$\Leftrightarrow x> \frac{-3}{2}$

d.

$\frac{x}{2}+3< 7$

$\Leftrightarrow \frac{x}{2}< 4$

$\Leftrightarrow x< 8$

b: \(\text{Δ}=\left(2m+2\right)^2-4\left(m^2+3m+2\right)\)

\(=4m^2+8m+4-4m^2-12m-8\)

=-4m-4

Để phương trình có hai nghiệm phân biệt thì -4m-4>0

=>-4m>4

hay m<-1

Theo đề, ta có: \(\left(x_1+x_2\right)^2-2x_1x_2=12\)

\(\Leftrightarrow\left(2m+2\right)^2-2\left(m^2+3m+2\right)-12=0\)

\(\Leftrightarrow4m^2+8m+4-2m^2-6m-4-12=0\)

\(\Leftrightarrow2m^2+2m-12=0\)

\(\Leftrightarrow m^2+m-6=0\)

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

=>m=-3(nhận) hoặc m=2(loại)

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

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

b) \(\left(x^2+5\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\\x^2-25=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=-5\\x^2=25\end{matrix}\right.\) \(\Leftrightarrow x^2=25\) \(\Leftrightarrow x=\pm5\)

a: =>x^3+2x^2-8x^2-16x+15x+30=0

=>(x+2)(x^2-8x+15)=0

=>(x+2)(x-3)(x-5)=0

=>\(x\in\left\{-2;3;5\right\}\)

b: =x^2-12x+36-3

=(x-6)^2-3>=-3

Dấu = xảy ra khi x=6