`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}

\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)

Lời giải:

a. $f'(x)\leq 0$

$\Leftrightarrow 3x^2-6x\leq 0$

$\Leftrightarrow x(x-2)\leq 0$

$\Leftrightarrow 0\leq x\leq 2$

b.

$f'(x)=x^2-3x+2=0$

$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$

$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

c.

$g(x)=f(1-2x)+x^2-x+2022$

$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$

$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$

$g'(x)\geq 0$

$\Leftrightarrow -24x^2+2x+5\geq 0$

$\Leftrightarrow (5-12x)(2x-1)\geq 0$

$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\text{ĐKXĐ:}x\ne3;-1;\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}MTC:2\left(x+1\right)\left(x-3\right)\)

\(\Rightarrow x^2+x+x^2-3x=4x\)

\(\Leftrightarrow2x^2-2x=4x\)

\(\Leftrightarrow2x^2-2x-4x=0\)

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\left(\text{loại}\right)\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{0\right\}\)

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne1;x\ne3\right)\\ \Leftrightarrow\dfrac{x.\left(x+1\right)+x.\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\\ \Rightarrow x^2+x+x^2-3x=4x\\ \Leftrightarrow2x^2-2x-4x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)

loại

Vậy phương trình có tập nghiệm S={\(\varnothing\)}

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 \(\left(x;y\right)=\left(3;3\right)\)

a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

 \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)

=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)

y'=0

=>\(\left(x+3\right)^2-4=0\)

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

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

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

=>\(\left(x-2\right)^2-9=0\)

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

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

=>x=5 hoặc x=-1

b.

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) hệ tương đương:

\(\left\{{}\begin{matrix}\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\\\dfrac{1}{x^3}+y^3=19\end{matrix}\right.\)

Đặt \(\left(\dfrac{1}{x};y\right)=\left(u;v\right)\) ta được: \(\left\{{}\begin{matrix}uv^2+u^2v=-6\\u^3+v^3=19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3uv^2+3u^2v=-18\\u^3+v^3+19\end{matrix}\right.\)

Cộng vế với vế:

\(\left(u+v\right)^3=1\Rightarrow u+v=1\)

Thay vào \(u^2v+uv^2=-6\Rightarrow uv=-6\)

Theo Viet đảo, u và v là nghiệm của:

\(t^2-t-6=0\) \(\Rightarrow\left[{}\begin{matrix}t=-2\\t=3\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(-2;3\right);\left(3;-2\right)\)

\(\Rightarrow\left(\dfrac{1}{x};y\right)=\left(-2;3\right);\left(3;-2\right)\)

\(\Rightarrow\left(x;y\right)=\left(-\dfrac{1}{2};3\right);\left(\dfrac{1}{3};-2\right)\)

a.

ĐKXĐ: \(x\ne3\)

- Với \(x\ge0\) pt trở thành:

\(\dfrac{x^2-x-12}{x-3}=2x\Rightarrow x^2-x-12=2x^2-6x\)

\(\Leftrightarrow x^2-5x+12=0\) (vô nghiệm)

- Với \(x< 0\) pt trở thành:

\(\dfrac{x^2+x-12}{x-3}=2x\Rightarrow\dfrac{\left(x-3\right)\left(x+4\right)}{x-3}=2x\)

\(\Rightarrow x+4=2x\Rightarrow x=4>0\) (ktm)

Vậy pt đã cho vô nghiệm

\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)

\(-\dfrac{2x}{\sqrt{2x-1}+2}>-\dfrac{1}{2};\dfrac{2x}{\sqrt{x+3}+4x}>0\)

Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm

Vậy PT có nghiệm duy nhất \(x=1\)

(x + 2)(x - 3)(x² + 2x - 24) = 16x² 

<=> (x + 2)(x - 3)[(x + 1)² - 25] = 16x² 

<=> (x + 2)(x - 3)(x + 6)(x - 4) = 16x² 

<=> (x² + 8x + 12)(x² - 7x + 12) = 16x² 

<=> \(\left(x^2+0,5x+12+7,5x\right)\left(x^2+0,5x+12-7.5x\right)=16x^2\)

<=> \(\left(x^2+0,5x+12\right)^2-\left(7,5x\right)^2=16x^2\)

<=> \(\left(x^2+0,5x+12\right)^2=\left(8,5x\right)^2\)

<=> \(\left(x^2+9x+12\right)\left(x^2-8x+12\right)=0\)

<=> \(\left(x+\dfrac{9}{2}-\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{9}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x-2\right)\left(x-6\right)=0\)

<=>\(\left[{}\begin{matrix}x=\dfrac{\sqrt{33}-9}{2}\\x=\dfrac{-\sqrt{33}-9}{2}\\x=2\\x=6\end{matrix}\right.\)