\(\Leftrightarrow\dfrac{4\cdot90\cdot\left(x+5\right)-4\cdot90\cdot x}{4x\left(x+5\right)}=\dfrac{x\left(x+5\right)}{4x\left(x+5\right)}\)

\(\Leftrightarrow x^2+5x-1800=0\)

\(\text{Δ}=5^2-4\cdot1\cdot\left(-1800\right)=7225>0\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-5-85}{2}=\dfrac{-90}{2}=-45\left(nhận\right)\\x_2=\dfrac{-5+85}{2}=40\left(nhận\right)\end{matrix}\right.\)

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)

=>(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

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<>0 và \(\begin{cases}x+\frac{1}{x^2}\ge0\\ x-\frac{1}{x^2}\ge0\end{cases}\)

=>\(\begin{cases}x^3+1\ge0\\ x^3-1\ge0\end{cases}\)

=>x>=1

\(\sqrt{x+\frac{1}{x^2}}+\sqrt{x-\frac{1}{x^2}}>\frac{2}{x}\)

=>\(\left( \sqrt{x + \frac{1}{x^2}} + \sqrt{x - \frac{1}{x^2}} \right)^2 > \left( \frac{2}{x} \right)^2\)

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

=>\(2x+2\sqrt{x^2 - \frac{1}{x^4}}>\frac{4}{x^2}\)

=>\(x + \sqrt{x^2 - \frac{1}{x^4}} > \frac{2}{x^2}\)

=>\(\sqrt{x^2 - \frac{1}{x^4}}>\frac{2}{x^2}-x\)

TH1: \(\frac{2}{x^2}-x\le0\)

=>\(\frac{2 - x^3}{x^2}\le0\)

=>\(x^3\ge2\)

=>\(x\ge\sqrt[3]{2}\)

Khi \(x\ge\sqrt[3]{2}\) thì VT<=0; VT>0

=>Bất phương trình luôn đúng với \(x\ge\sqrt[3]{2}\) (2)

TH2: \(\frac{2}{x^2}-x>0\)

=>\(\frac{2 - x^3}{x^2}>0\)

=>\(x^3<2\)

=>\(x<\sqrt[3]{2}\)

BPT sẽ tương đương: \(x^2 - \frac{1}{x^4} > \left( \frac{2}{x^2} - x \right)^2\)

=>\(x^2 - \frac{1}{x^4} > \frac{4}{x^4} - \frac{4}{x} + x^2\)

=>\(-\frac{1}{x^4} > \frac{4}{x^4} - \frac{4}{x}\)

=>\(\frac{4}{x} > \frac{5}{x^4}\)

=>\(4x^3>5\)

=>\(x^3>\frac54=\frac{10}{8}\)

=>\(x>\frac{\sqrt[3]{10}}{2}\)

=>\(\frac{\sqrt[3]{10}}{2} (1)

Từ (1),(2) suy ra \(S = \left( \sqrt[3]{\frac{5}{4}}; +\infty \right)\)

Ta có: \(\dfrac{1}{x^2} + \dfrac{x^2}{1-x^2} + \dfrac{5}{2}\left(\dfrac{\sqrt{1-x^2}}{x} + \dfrac{x}{\sqrt{1-x^2}}\right) + 2 > 0\) (1)

ĐKXĐ: \(\begin{cases}1-x^2>0\\ x<>0\end{cases}\Rightarrow\begin{cases}x^2<1\\ x<>0\end{cases}\)

=>-1<x<1 và x<>0

Đặt \(y=\dfrac{\sqrt{1-x^2}}{x}+\dfrac{x}{\sqrt{1-x^2}}\)

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

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

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

Thay vào BPT ban đầu, ta được: \((y^2 - 1) + \dfrac{5}{2}y + 2 > 0\)

=>\(y^2+\dfrac{5}{2}y+1>0\iff2y^2+5y+2>0\)

=>(2y+1)(y+2)>0

=>y>-1/2 hoặc y<-2

TH1: 0<x<1

Theo AM-GM, ta được: \(y \ge 2 \cdot \sqrt{\dfrac{\sqrt{1-x^2}}{x} \cdot \dfrac{x}{\sqrt{1-x^2}}} = 2\)

Nếu y>-1/2 thì vì y>=2>-1/2

nên (1) luôn đúng

Nếu y<-2 thì vì y>=2>-2

nên (1) vô nghiệm

TH2: -1<x<0

=>\(\frac{\sqrt{1-x^2}}{x}<0;\frac{x}{\sqrt{1-x^2}}<0\)

Đặt \(t=-\dfrac{\sqrt{1-x^2}}{x}\)

=>\(y=-t-\dfrac{1}{t}=-\left(t+\dfrac{1}{t}\right)\le-2\)

Nếu y>-1/2 thì vì y<=-2<-1/2

nên (1) luôn sai

=>(1) vô nghiệm

Nếu y<-2 thì vì y<=-2<-2

nên mọi giá trị của x sẽ thỏa mãn y, trừ giá trị thỏa mãn y=-2

Đặt y=-2

=>\(t+\dfrac{1}{t}=2\iff t=1\)

=>\(-\dfrac{\sqrt{1-x^2}}{x}=1\)

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

=>\(1-x^2=x^2\)

=>\(2x^2=1\)

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

Do đó: x∈\(\left(-1;\frac{-1}{\sqrt2}\right)\) \(\cup\) \(\left(-\frac{1}{\sqrt2};0\right)\)

Vậy: \(S = \left(-1, -\dfrac{1}{\sqrt{2}}\right) \cup \left(-\dfrac{1}{\sqrt{2}}, 0\right) \cup (0, 1)\)

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

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

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

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

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

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

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

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

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

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

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

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow x^2+x+2x+2-5x+10=12+x^2-4\\ \Leftrightarrow-2x=-4\\ \Leftrightarrow x=2\left(ktm\right)\)

Vậy \(S\in\left\{\varnothing\right\}\)

ĐKXĐ: \(\begin{cases}x-2\ne 0\\x+2\ne 0\end{cases}\leftrightarrow x\ne 2\\x\ne -2\end{cases}\)

\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\leftrightarrow \dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\dfrac{5(x-2)}{(x+2)(x-2)}=\dfrac{12}{(x-2)(x+2)}+\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\)

\(\to x^2+3x+2-5x+10=12+x^2-4\)

\(\leftrightarrow x^2-2x-x^2=12-12-4\)

\(\leftrightarrow -2x=-4\)

\(\leftrightarrow x=2(\rm KTM)\)

Vậy pt đã cho vô nghiệm \(S=\varnothing\)

ĐKXĐ : \(x\notin\left\{0;-1;-2;-3;-4\right\}\)

Ta có \(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)

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

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

Đặt x + 2 = a \(\left(a\ne0\right)\) 

(*) \(\Leftrightarrow\dfrac{2a}{a^2-4}+\dfrac{2a}{a^2-1}+\dfrac{1}{a}=0\)

\(\Leftrightarrow\dfrac{2}{a-\dfrac{4}{a}}+\dfrac{2}{a-\dfrac{1}{a}}+\dfrac{1}{a}=0\) (**)

Đặt \(\dfrac{1}{a}=b\left(b\ne0\right)\) \(\Rightarrow ab=1\)

Ta được (**) \(\Leftrightarrow\dfrac{2}{a-4b}+\dfrac{2}{a-b}+b=0\)

\(\Leftrightarrow\dfrac{2b}{1-4b^2}+\dfrac{2b}{1-b^2}+b=0\)

\(\Leftrightarrow\dfrac{2}{1-4b^2}+\dfrac{2}{1-b^2}=-1\)

\(\Rightarrow4-10b^2=-4b^4+5b^2-1\)

\(\Leftrightarrow4b^4-15b^2+5=0\) (***)

Đặt b² = t > 0

Ta có (***) <=> \(4t^2-15t+5=0\Leftrightarrow t=\dfrac{15\pm\sqrt{145}}{8}\) (tm)

\(\Leftrightarrow b=\pm\sqrt{\dfrac{15\pm\sqrt{145}}{8}}\) 

mà x + 2 = a ; ab = 1 

nên \(x=\pm\sqrt{\dfrac{8}{15\pm\sqrt{145}}}-2\)

Thử lại ta có phương trình có 4 nghiệm như trên