`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}

TH1: \(x\ge2\)

\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=4\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=4\)

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

TH2: \(x< 2\)

\(-\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=-4\)

\(\Leftrightarrow x^4-5x^2+8=0\)

\(\Leftrightarrow\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\) (vô nghiệm)

Vậy \(x=\sqrt{5}\)

Bài 4 :

24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ

Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0 

Suy ra quãng đường AB là 36x(km)

Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)

Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)

Ta có phương trình: 

\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)

Vậy quãng đường AB dài 36.2 = 72(km)

Bài 3 : 

\(A = -x^2 + 2x + 9 = -(x^2 -2x - 9) \\= -(x^2 - 2x + 1 + 10) = -(x^2 -2x + 1)+ 10\\=-(x-1)^2 + 10\)

Vì : \((x-1)^2 \geq 0\) ∀x \(\Leftrightarrow -(x-1)^2 \)≤ 0 ∀x \(\Leftrightarrow -(x-1)^2 + 10\) ≤ 10

Dấu "=" xảy ra khi và chỉ khi x - 1 = 0 ⇔ x = 1

Vậy giá trị nhỏ nhất của A là 10 khi x = 1

\(ĐK:x\ne0;x\ne1\\ PT\Leftrightarrow\left(\dfrac{1}{x}+2\right)\left(2+\dfrac{x+1}{x-1}-x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{x}=-2\\\dfrac{x+1}{x-1}=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x+1=x^2-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x^2-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)

\(\left(x+1\right)^2+\left|x-7\right|+6=\left(x+2\right)^2\)

\(< =>x^2+2x+1+\left|x-7\right|+6=x^2+4x+4\)

\(< =>\left|x-7\right|=x^2-x^2+4x-2x+4-1-6\)

\(< =>\left|x-7\right|=2x-3\)

\(< =>\left[{}\begin{matrix}x-7=2x-3\\x-7=-2x+3\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x-2x=-3+7\\x+2x=3+7\end{matrix}\right.\\ < =>\left[{}\begin{matrix}-x=4\\3x=10\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{3}\end{matrix}\right.\)

\(ĐK:x\ne-2\\ PT\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2+2\right)}{x+2}=4\\ \Leftrightarrow\left(x+1\right)\left(x+4\right)=4\left(x+2\right)\\ \Leftrightarrow x^2+5x+4=4x+8\\ \Leftrightarrow x^2+x-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{17}}{2}\\x=\dfrac{-1-\sqrt{17}}{2}\end{matrix}\right.\)

TH1: \(x\le-2\)

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

\(\Leftrightarrow0x=5\)

\(\Rightarrow\) vô nghiệm

TH2: \(-2< x\le-1\)

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

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

TH3: \(-1< x\le\dfrac{1}{2}\)

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

\(\Leftrightarrow0x=-1\)

\(\Rightarrow\) vô nghiệm

TH4: \(x>\dfrac{1}{2}\)

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

\(\Leftrightarrow x=\dfrac{1}{4}\left(l\right)\)

Vậy phương trình đã cho vô nghiệm

${}^{}+{}^{}={}^{}\iff {}^{}+3{}^{}+3x+1+{}^{}-6{}^{}+12x-8=8{}^{}-12{}^{}+6x-1\iff 2{}^{}-3{}^{}+15x-7-8{}^{}+12{}^{}-6x+1=0$$\iff -6{}^{}$

Đặt x+1=a; x-2=b

Phương trình trở thành:

\(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

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

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

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=8x^3-12x^2+6x-1\)

\(\Leftrightarrow2x^3-3x^2+15x-7-8x^3+12x^2-6x+1=0\)

\(\Leftrightarrow-6x^3+9x^2+9x-6=0\)

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

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

\(\Leftrightarrow2\left(x^3+1\right)-3x\left(x+1\right)=0\)

\(\Leftrightarrow2\left(x^2-x+1\right)\left(x+1\right)-3x\left(x+1\right)=0\)

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

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

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

\(\Leftrightarrow\left[\left(2x^2-4x\right)-\left(x-2\right)\right]\left(x+1\right)=0\)

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

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

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

\(\text{Vậy tập nghiệm phương trình là:}\left\{\dfrac{1}{2};2;\left(-1\right)\right\}\)

\(ĐK:x\ne-1\\ PT\Leftrightarrow\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2+x+1\right)}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{1}{4}\\ \Leftrightarrow4x-4=x+1\\ \Leftrightarrow3x=5\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)

Ta xét các trường hợp :

TH1 : \(x< 1\Rightarrow x-1,x-2< 0\Rightarrow\left|x-1\right|=1-x;\left|x-2\right|=2-x\)

\(\Rightarrow1-x+2-x=5\Leftrightarrow2x=-2\Leftrightarrow x=-1\left(t.m\right)\)

TH2 : \(1\le x< 2\Rightarrow x-2< 0\le x-1\Rightarrow\left|x-1\right|=x-1;\left|x-2\right|=2-x\\ \Rightarrow x-1+2-x=5\Leftrightarrow1=5\left(VL\right)\)

TH3: \(x\ge2\Rightarrow x-1,x-2\ge0\Leftrightarrow\left|x-1\right|=x-1;\left|x-2\right|=x-2\)

\(\Rightarrow x-1+x-2=5\Leftrightarrow2x=8\Leftrightarrow x=4\left(t.m\right)\)