√3/2 = cos30o nên cos⁡(x + 30o )= √3/2

⇔ cos⁡(x + 30o ) = cos 30o

⇔ x + 30o = ±30o + k360o, k ∈ Z

⇔ x = k360o, k ∈ Z và x = -60o + k360o, k ∈ Z

đk : x khác 1 ; -1 

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

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

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

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

=>-3x+4=0

hay x=4/3(nhận)

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

\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\left(quy\cdotđồng\cdot và\cdot khử\cdot mẫu\right)\)

\(\Leftrightarrow-3x+4=0\)

\(\Leftrightarrow-3x=-4\)

\(\Leftrightarrow x=\dfrac{-4}{3}\)

Vậy \(S=\left\{-\dfrac{4}{3}\right\}\)

=>(x^2+1)^2+x^2/x*(x^2+1)=5/2

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

=>\(2\left(x^4+2x^2+1+x^2\right)=5\left(x^3+x\right)\)

=>2x^4+6x^2+2-5x^3-5x=0

=>2x^4-5x^3+6x^2-5x+2=0

=>2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0

=>(x-1)(2x^3-3x^2+3x-2)=0

=>(x-1)(2x^3-2x^2-x^2+x+2x-2)=0

=>(x-1)^2*(2x^2-x+2)=0

=>x-1=0

=>x=1

2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

1, ĐKXĐ:\(x\ne2,y\ne1\)

Đặt `1/(x-2)` = a, `1/(y-1)` = b

\(Hệ.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\\b=\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{y-1}=\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\3y-3=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\\y=\dfrac{8}{3}\end{matrix}\right.\)\(2,\Delta'=\left[-\left(m+1\right)\right]^2-4m=m^2+2m+1-4m=m^2-2m+1=\left(m-1\right)^2\ge0\)

Để pt có 2 nghiệm phân biệt thì \(\Delta'>0\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m-1\ne0\Leftrightarrow m\ne1\)

b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=4m\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2-x_1x_2=3\\ \Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2=3\\ \Leftrightarrow\left(2m+2\right)^2-5.4m-3=0\\ \Leftrightarrow4m^2+8m+4-20m-3=0\\ \Leftrightarrow4m^2-12m+1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\)

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

\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow x\left(x+8\right)=105\)

\(\Leftrightarrow x^2+8x-105=0\)

\(\Leftrightarrow x^2-7x+15x-105=0\)

\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)

Sửa đề: \(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)

ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

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

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

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

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

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

=>\(x^2-x=0\)

=>x(x-1)=0

=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

ĐKXĐ : \(x\ne\pm2\)

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

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

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

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

\(\Leftrightarrow x=\dfrac{-3\pm\sqrt{5}}{2}\left(TM\right)\)

Vậy ...