`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.

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

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

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

$g'(x)\geq 0$

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

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

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

c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

\(\Leftrightarrow\left(m+1\right)\left(2m-1\right)\le0\)

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)

\(x^2-2x-2-2\sqrt{2x+1}=0\)

\(\Leftrightarrow x^2-2x-8-\left(2\sqrt{2x+1}-6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)-\frac{4\left(2x+1\right)-36}{2\sqrt{2x+1}+6}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)-\frac{8\left(x-4\right)}{2\sqrt{2x+1}+6}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2-\frac{8}{2\sqrt{2x+1}+6}\right)=0\)

Thấy: \(x+2-\frac{8}{2\sqrt{2x+1}+6}>0\)

\(\Rightarrow x-4=0\Rightarrow x=4\)

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\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24=1.2.3.4=\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

\(S=\left\{-2;0\right\}\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+4=a\)

\(pt\Leftrightarrow a\left(a+2\right)-24=0\)

\(\Leftrightarrow a^2+2a-24=0\)

\(\Leftrightarrow a^2+6a-4a-24=0\)

\(\Leftrightarrow a\left(a+6\right)-4\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right)\left(a-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=4\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2+5x+4=-6\\x^2+5x+4=4\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x+10=0\\x^2+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\\x\left(x+5\right)=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{5}{2}\right)^2=\frac{-15}{4}\left(loai\right)\\x\in\left\{0;-5\right\}\end{cases}}\)

Vậy....

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

Trường hợp 1: x<=-2

=>-x-2-x+3=5

=>-2x+1=5

=>-2x=4

hay x=-2(nhận)

Trường hợp 2: -2<x<3

=>x+2+3-x=5

=>5=5(đúng)

Trường hợp 3: x>=3

=>x+2+x-3=5

=>2x-1=5

=>2x=6

hay x=3(nhận)

Vậy: -2<=x<=3

\(\Leftrightarrow x+2+x-3=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

x+2+x−3=5⇔2x=6⇔x=3

x+2+x−3=5⇔2x=6⇔x=3

Trường hợp 1: x<=-2

=>-x-2-x+3=5

=>-2x+1=5

=>-2x=4

hay x=-2(nhận)

Trường hợp 2: -2<x<3

=>x+2+3-x=5

=>5=5(đúng)

Trường hợp 3: x>=3

=>x+2+x-3=5

=>2x-1=5

=>2x=6

hay x=3(nhận)

Vậy: -2<=x<=3

`(x-5)^2 +3(x-5)=0`

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

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

\(< =>\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

PT \(\Leftrightarrow\left[x\left(x+1\right)\right].\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(x^2+x=t\) ta được:

\(t\left(t-2\right)-24=0\Leftrightarrow t^2-2t-24=0\)

\(\Leftrightarrow t^2-6t+4t-24=0\)\(\Leftrightarrow t\left(t-6\right)+4\left(t-6\right)=0\)

\(\Leftrightarrow\left(t-6\right)\left(t+4\right)=0\Leftrightarrow\orbr{\begin{cases}t=6\\t=-4\end{cases}}\)

Suy ra \(\orbr{\begin{cases}x^2+x-6=0\\x^2+x+4=0\end{cases}}\)

Ez rồi.