`x(x+3) - (2x-1) . (x+3) = 0`

`<=>(x+3)(x-2x+1)=0`

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

`** x+3=0`

`<=>x=-3`

`** -x+1=0`

`<=>x=1`

`x(x-3) - 5 (x-3) = 0`

`<=>(x-3)(x-5)=0`

`** x-3=0`

`<=>x=3`

`** x-5=0`

`<=>x=5`

`3x + 12 = 0`

`<=>3x=-12`

`<=> x=-4`

`2x (x-2) + 5 (x-2) = 0`

`<=>(x-2)(2x+5)=0`

`** x-2=0`

`<=>x=2`

`** 2x+5=0`

`<=> x= -5/2`

\(x+x^2=0\)

\(\Leftrightarrow x\left(1+x\right)=0\)

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

Vậy: Phương trình có tập nghiệm \(S=\left\{0;-1\right\}\)

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

Tham khảo thử đúng không nha mn

     \(x^2+x-y^2=0\)

⇔ \(\left(x^2-y^2\right)+x=0\)

⇔ \(\left(x-y\right)\left(x+y\right)+x=0\)

⇒ \(x-y=0\) hoặc \(x+y=0\) hoặc \(x=0\)

⇒ \(x=y=0\)

\(\Leftrightarrow x^2+x=y^2\)

\(\Leftrightarrow4x^2+4x=4y^2\)

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

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

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

Vậy \(\left(x;y\right)=\left(0;0\right);\left(-1;0\right)\)

ta có x³-6x²+11x-6=0

hay x³-x²-5x²-+5x+6x-6=0

=>x(x-1) - 5x(x-1)+6(x-1)=0

(x-1).(x-5x+6)=0 <=> (x-1)(x²-2x-3x+6)=0

(x-1)(x(x-2)-3(x-2)=0

(x-1)(x-2)(x-3)=0 <=> x-1=0 hoặc x-2=0 hoặc x-3=0

<=> x=1 hoặc x=2 hoặc x=3

vậy S ={1;2;3}

`x(x - 4) - 3x + 12 = 0`

`<=> x(x - 4) + 3(x - 4) = 0`

`<=> (x + 3)(x - 4) = 0`

`<=>` $\left[\begin{matrix} x + 3 = 0\\ x - 4 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -3\\ x = 4\end{matrix}\right.$

Vậy `S = {-3; 4}`

\(x\left(x-4\right)-3x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

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

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

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

=> 3x-(1/2013+2/2012+3/2011)=3x-(4/2010+5/2009+6/2008)=>6x=-4/2010-5/2009-6/2008+1/2013+2/2012+3/2011                                                                                       =>x=...                                                                          làm tiếp đi bạn

\(\left\{{}\begin{matrix}x_1+x_2=3\\A=x_1x_2=-7\end{matrix}\right.\)

\(B=\left(x_1+x_2\right)^2-2x_1x_2=9-2\left(-7\right)=9+14=23\)

Bài làm:

a) \(4x\left(x+2\right)=4x^2-24\)

\(\Leftrightarrow4x^2+8x=4x^2-24\)

\(\Leftrightarrow8x=-24\)

\(\Leftrightarrow x=-3\)

Vậy tập nghiệm của phương trình \(S=\left\{-3\right\}\)

b) \(\frac{x-2}{3}< \frac{8x-5}{9}\)

\(\Leftrightarrow\frac{3\left(x-2\right)}{9}< \frac{8x-5}{9}\)

\(\Leftrightarrow3x-6< 8x-5\)

\(\Leftrightarrow-5x< 1\)

\(\Leftrightarrow x>-\frac{1}{5}\)

Vậy \(x>-\frac{1}{5}\)

c) đkxđ: \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

Ta có: \(\frac{3}{x-2}+\frac{2}{x+2}=\frac{2x+5}{x^2-4}\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+5}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow3\left(x+2\right)+2\left(x-2\right)=2x+5\)

\(\Leftrightarrow3x+6+2x-4=2x+5\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy tập nghiệm của phương trình \(S=\left\{1\right\}\)

Học tốt!!!!