\(3.\frac{x}{x^2+1}=1\)

\(\Leftrightarrow\frac{x}{x^2+1}=\frac{1}{3}\)

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

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

\(\cdot\Delta=\left(-3\right)^2-4.1.1=5\)

\(\cdot\Delta>0\)nên pt có 2 nghiệm phân biệt

\(x_1=\frac{3+\sqrt{5}}{2}\);\(x_2=\frac{3-\sqrt{5}}{2}\)

\(\Delta\)là gì ạ?

\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

b.\(Q< 1\)

\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)

\(\Leftrightarrow4\sqrt{x}-8< 0\)

\(\Leftrightarrow0\le x< 4\)

Vay de Q<1 thi \(0\le0< 4\)

\(A=\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}=x\left(1-\frac{y^2}{1+y^2}\right)+y\left(1-\frac{z^2}{1+z^2}\right)+z\left(1-\frac{x^2}{1+x^2}\right)\)

\(\Rightarrow A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)=\left(x+y+z\right)-\frac{xy+yz+zx}{2}\ge3-\frac{\frac{9}{3}}{2}=\frac{3}{2}\)

Dau '=' xay ra khi \(x=y=z=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(x=y=z=1\)

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow\)\(a+b+c\ge3\left(\frac{ab+bc+ca}{a+b+c}\right)\)

\(\Leftrightarrow\)\(a+b+c\ge3\left(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\right)\)

\(\Leftrightarrow\)\(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)

\(\sqrt{3x^2+2}=t\left(t\ge0\right)\)

\(pt\Leftrightarrow t^2-7t+4=0\)

\(\Leftrightarrow t=\frac{7\pm\sqrt{33}}{2}\)

bạn giải típ nhá

\(pt\Leftrightarrow2x^2-17+20-5\sqrt{2x^2-1}=0\)

\(\Leftrightarrow2x^2-17+\frac{5\cdot\left(17-2x^2\right)}{20+5\sqrt{2x^2-1}}=0\)

\(\Leftrightarrow\left(2x^2-17\right)\left(1-\frac{5}{20+5\sqrt{2x^2-1}}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{17}{2}}\\20+5\sqrt{2x^2-1}=5\end{cases}}\)\(\Leftrightarrow5\sqrt{2x^2-1}=-15\)(vô lý)

Vậy \(x=\sqrt{\frac{17}{2}}\)