ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)

Ta có: \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

Suy ra: \(x^2+5x+4=18\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow x^2+7x-2x-14=0\)

\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-7;2}

ĐKXĐ: $x \neq -1;-2;-3;-4$

$pt⇔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}$

$⇔\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}$

$⇔\dfrac{3}{(x+1)(x+4)}=\dfrac{1}{6}$

$⇔x^2+5x+4=18$

$⇔x^2+5x-14=0$

$⇔(x-2)(x+7)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)(t/m)

Vậy...

PT tương đương

\(\left(x^2+7x+6\right)\left(x^2+5x+6\right)=\dfrac{-3x^2}{4}\)

Xét \(x=0\Rightarrow6.6=0\)(vô lý)

Xét \(x\ne0\). Ta chia 2 vế của PT cho \(x^2\ne0\). PT tương đương

\(\left(x+\dfrac{6}{x}+7\right)\left(x+\dfrac{6}{x}+5\right)=\dfrac{-3}{4}\)

Đặt \(x+\dfrac{6}{x}+5=t\)

PT\(\Leftrightarrow t\left(t+2\right)=\dfrac{-3}{4}\Leftrightarrow t^2+2t+1=\dfrac{1}{4}\)

\(\Leftrightarrow\left(t+1\right)^2=\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}t+1=\dfrac{-1}{2}\\t+1=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-3}{2}\\t=\dfrac{-1}{2}\end{matrix}\right.\)

Đến đây bạn thay vào là tìm được nghiệm nhé.

`(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)`

`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

<=>

ĐK: \(x\ge1\)

\(pt\Leftrightarrow2\sqrt{\left(x-1\right)\left(x+2\right)}-\sqrt{x-1}-6\sqrt{x+2}+3=0\)

\(\Leftrightarrow\left(2\sqrt{x+2}-1\right)\left(\sqrt{x-1}-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+2\right)=1\\x-1=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{4}\left(l\right)\\x=10\left(tm\right)\end{matrix}\right.\)

Vậy ...

1) \(2\left(x+3\right)>5\left(x-1\right)+2\Leftrightarrow2x+6>5x-5+2\Leftrightarrow3x>9\Leftrightarrow x>3\)

2) \(x^2-x\left(x+2\right)>3x-10\)

\(\Leftrightarrow x^2-x^2-2x>3x-10\Leftrightarrow5x< 10\Leftrightarrow x< 2\)

3) \(x\left(x-5\right)< \left(x+1\right)^2\)

\(\Leftrightarrow x^2-5x< x^2+2x+1\Leftrightarrow7x>-1\Leftrightarrow x>-\dfrac{1}{7}\)

4) \(15-2\left(x-7\right)< 2\left(x-3\right)-6\)

\(\Leftrightarrow15-2x+14< 2x-6-6\Leftrightarrow4x>41\Leftrightarrow x>\dfrac{41}{4}\)

1: Ta có: \(2\left(x+3\right)>5\left(x-1\right)+2\)

\(\Leftrightarrow2x+6>5x-5+2\)

\(\Leftrightarrow-3x>-9\)

hay x<3

2: Ta có: \(x^2-x\left(x+2\right)>3x-10\)

\(\Leftrightarrow x^2-x^2-2x>3x-10\)

\(\Leftrightarrow-5x>-10\)

hay x<2

3: Ta có: \(x\left(x-5\right)\le\left(x+1\right)^2\)

\(\Leftrightarrow x^2-5x-x^2-2x-1\ge0\)

\(\Leftrightarrow-7x\ge1\)

hay \(x\le-\dfrac{1}{7}\)

4: Ta có: \(15-2\left(x-7\right)< 2\left(x-3\right)-6\)

\(\Leftrightarrow15-2x+14< 2x-6-6\)

\(\Leftrightarrow-4x< -12-29=-41\)

hay \(x>\dfrac{41}{4}\)

\(1,\) thiếu đề

\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-55x+20=24x-138\)

\(\Leftrightarrow24x-138+55x-20=0\)

\(\Leftrightarrow79x-158=0\)

\(\Leftrightarrow x=2\)

\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)

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

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

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

\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

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

\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

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

\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)

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

\(\Leftrightarrow x=0\)

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

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.

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

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

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

$g'(x)\geq 0$

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

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

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