a)

∆'=3+9=12

x1=(√3-2√3)/3=-√3/3

x2=(√3+2√3)/3=√3

b.

<=>

x+y=-3(1)

2x-3y=-1(2)

(1).2-(2)<=>5y=-5;y=-1

=>(x,y)=(-2;-1)

nếu là lớp 8 thì rất hoan nghênh

a) \(\Delta'=\left(\sqrt{3}\right)^2-3\cdot\left(-3\right)=12>0\)

phương trình có 2 nghiệm phân biệt:

\(\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+\sqrt{12}}{3}=\sqrt{3}\\x=\dfrac{\sqrt{3}-\sqrt{12}}{3}=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

kết luận: \(x=\sqrt{3}\), \(x=-\dfrac{\sqrt{3}}{3}\)

b) \(\left\{{}\begin{matrix}x\left(x-1\right)+y=\left(x+1\right)\left(x-3\right)\\2x-3y=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\\5x+9=-1\end{matrix}\right.\)

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

kết luận: \(\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

a) lập delta giải bình thường

b) rút y từ 1 trong 2 pt thế vào pt còn lại

@Leo :) giỏi wa ha . ghen tị ghê

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

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

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

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

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

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

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

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

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

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

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

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

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

Vậy: S={3;-3}

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. 

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x=2x^3-16\)

<=>\(8x=-16\)

<=>\(x=-2\)

i. 

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>\(2x^2-x-3=2x^2+9x-5\)

<=>10x=2

<=>\(x=\dfrac{1}{5}\)

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

<=>

<=>

<=>\(8x=-16\)

<=>x=-2

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

<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(9x+6=0\)

<=>x=\(\dfrac{-2}{3}\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>

<=>10x=2

<=>

Không hiểu câu hỏi lắm :vvv