1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

\(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)

\(\Leftrightarrow\dfrac{2x}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\)

\(ĐK:x\ne1;-3\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)+4}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)

\(\Leftrightarrow13x=1\)

\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

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

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

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

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

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

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

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

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

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

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

a)Pt\(\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=x+\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)

\(\Leftrightarrow x+\sqrt{3}\ge0\)\(\Leftrightarrow x\ge-\sqrt{3}\)

Vậy...

b)Đk:\(x\ge4\)

Pt\(\Leftrightarrow\sqrt{\left(x-4\right)+2\sqrt{x-4}+1}=2\sqrt{x-4}+1\)

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

\(\Leftrightarrow\sqrt{x-4}+1=2\sqrt{x-4}+1\)

\(\Leftrightarrow\sqrt{x-4}=0\)

\(\Leftrightarrow x=4\) (tm)

Vậy...

a) Ta có: \(\sqrt{x^2+2x\sqrt{3}+3}=x+\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x\ge-\sqrt{3}\\x=-\sqrt{3}\left(loại\right)\end{matrix}\right.\Leftrightarrow x\ge-\sqrt{3}\)

a) \(\left(x-1\right)^3\)

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

b) \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

Bài 3: 

a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)

\(\Leftrightarrow12x=13\)

hay \(x=\dfrac{13}{12}\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)

\(\Leftrightarrow x^3-1-x^3+4x=4\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

Bài 2:

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

  Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

  Vậy \(A_{Min}=-2\) khi \(x=-1\)

Bài 1:

a) Ta có: \(2x^2-6=0\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)

b, \(đk:x\ge2\)

Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0

 \(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)

\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)

\(\Leftrightarrow x^3-11x^2+35x-25\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\)  (*)

Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)

Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5

c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)

\(\Leftrightarrow4x^3+x>0\)

Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))

\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)

\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....

d) Đk: \(x\ge\dfrac{3}{4}\)

Áp dụng bđt cosi:

 \(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)

 \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)

\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)

\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)

Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)

Dấu = xảy ra khi x=1 (tm)

`a)\sqrtx+\sqrt{2-x}=(3x^2-2x+3)/(x^2+1)`

`đk:0<=x<=2`

`pt<=>sqrtx-1+\sqrt{2-x}-1=(3x^2-2x+3)/(x^2+1)-2`

`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x^2-2x+1)/(x^2+1)`

`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x-1)^2/(x^2+1)`

`<=>(x-1)((x-1)/(x^2+1)+1/(sqrt{2-x}+1)-1/(sqrtx+1))=0`

`<=>x-1=0<=>x=1`

Vậy `S={1}`

`2/(4-x^2)+1/(x^2-2x)=(x-4)/(x^2+2x)(x ne 0,+-2)`

`<=>(2x)/(4x-x^3)+(x+2)/(x^3-4x)=(x^2-6x+8)/(x^3-4x)`

`<=>-2x+x+2=x^2-6x+8`

`<=>x^2-7x+10=0`

`<=>x^2-2x-5x+10=0`

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

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

Vì `x ne 2=>x-2 ne 0`

`=>x-5=0`

`=>x=5`

Vậy `S={5}`

b) ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\)

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

Suy ra: \(2x^2+2x+1-3x^2-x^2+x=0\)

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

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

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

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

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

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)