`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}

${}^{}+{}^{}={}^{}\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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1) đkxđ \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\y\ge0\end{matrix}\right.\)

Xét biểu thức \(P=x^3+y^3+7xy\left(x+y\right)\)

\(P=\left(x+y\right)^3+4xy\left(x+y\right)\)

\(P\ge4\sqrt{xy}\left(x+y\right)^2\)

Ta sẽ chứng minh \(4\sqrt{xy}\left(x+y\right)^2\ge8xy\sqrt{2\left(x^2+y^2\right)}\)  (*)

Thật vậy, (*)

\(\Leftrightarrow\left(x+y\right)^2\ge2\sqrt{2xy\left(x^2+y^2\right)}\)

\(\Leftrightarrow\left(x+y\right)^4\ge8xy\left(x^2+y^2\right)\)

\(\Leftrightarrow x^4+y^4+6x^2y^2\ge4xy\left(x^2+y^2\right)\) (**)

Áp dụng BĐT Cô-si, ta được:

VT(**) \(=\left(x^2+y^2\right)^2+4x^2y^2\ge4xy\left(x^2+y^2\right)\)\(=\) VP(**)

Vậy (**) đúng \(\Rightarrowđpcm\). Do đó, để đẳng thức xảy ra thì \(x=y\). 

Thế vào pt đầu tiên, ta được \(\sqrt{2x-3}-\sqrt{x}=2x-6\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{2x-3}+\sqrt{x}}=2\left(x-3\right)\)

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

 Rõ ràng với \(x\ge\dfrac{3}{2}\) thì \(\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}\le\dfrac{1}{\sqrt{\dfrac{2.3}{2}-3}+\sqrt{\dfrac{3}{2}}}< 2\) nên ta chỉ xét TH \(x=3\Rightarrow y=3\) (nhận)

Vậy hệ pt đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(3;3\right)\)

3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36

=> 3(x^2 + 4x + 4) + 4x^2 - 4x + 1 - 7(x^2 - 9) = 36

=> 3x^2 + 12x + 12 + 4x^2 - 4x + 1 - 7x^2 + 63 = 36

=> 8x + 76 = 36

=> 8x = -40

=> x = -5

a) 16x² - ( x² + 4 )²

= ( 4x )² - ( x² + 4 )²

= [ 4x - ( x² + 4 ) ][ 4x + ( x² + 4 ) ]

= ( -x² + 4x - 4 )( x² + 4x + 4 )

= [ -( x² - 4x + 4 ) ]( x + 2 )²

= [ -( x - 2 )² ]( x + 2 )²

b) ( x + y )³ + ( x - y )³

= [ ( x + y ) + ( x - y ) ][ ( x + y )² - ( x + y )( x - y ) + ( x - y )² ]

= ( x + y + x - y )[ x² + 2xy + y² - ( x² - y² ) + x² - 2xy + y² ]

= 2x( 2x² + 2y² - x² + y²

= 2x( x² + 3y² )

\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)

Mà \(-3< 0.\)

\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)

\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)

Kết hợp ĐKXĐ.

\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)

\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)

Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)

Ta có bảng sau:

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].

2) −(x2−2x+5)x−2−x+1≥0−(x2−2x+5)x−2−x+1≥0                                                        ⇔$-{}^{}$

Lời giải:
ĐK: $x,y,z\geq 0$

Áp dụng BĐT Cô-si:

\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\geq 3\sqrt[3]{\frac{xyz}{(x+1)(y+1)(z+1)}}\)

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq 3\sqrt[3]{\frac{1}{(x+1)(y+1)(z+1)}}\)

Cộng theo vế và thu gọn:

\(3\geq 3.\frac{\sqrt[3]{xyz}+1}{\sqrt[3]{(x+1)(y+1)(z+1)}}\Leftrightarrow (x+1)(y+1)(z+1)\geq (1+\sqrt[3]{xyz})^3\)

Dấu "=" xảy ra khi $x=y=z$

Thay vào pt $(1)$ thì suy ra $x=y=z=1$

`(x+1)^2 -|3-2x| +6 = (x+2)^2`

`<=> x^2 +2x +1 -|3-2x| +6 = x^2 +4x +4`

`<=> 2x +7 -4x -4 -|3-2x| =0`

`<=> 3 -2x -|3-2x| =0`

`<=> |3-2x| = 3-2x`

`@` nếu` 3-2x >= 0 => x <= 3/2  => |3-2x| =3-2x`

`=>` PT có dạng 

`3-2x =3-2x(luôn-đúng)`

`=>` PT luôn có nghiệm khi `x<=3/2`

`@` nếu` 3-2x <0 => x >3/2 => |3-2x| = 2x-3`

`=> PT có dạng 

`2x-3 = 3-2x`

`<=> 2x +2x = 3+3`

`<=> 4x=6`

`=> x = 3/2( loại)`