điều kiện : \(x>\frac{1}{3}\)

Phương trình \(\Leftrightarrow\frac{3x+1}{\sqrt{3x-1}}=\sqrt{5x+3}\Leftrightarrow3x+1=\sqrt{\left(3x-1\right)\left(5x+3\right)}\)

\(\Leftrightarrow9x^2+6x+1=15x^2+4x-3\Leftrightarrow6x^2-2x-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{2}{3}\text{ loại}\end{cases}}\) vậy x=1

 alibaba nguyễn giúp em

Áp dụng BĐT Bunhiacopxki

\(\left(a+b\right)\left(\frac{4}{a}+\frac{1}{4b}\right)\ge\left(\sqrt{a}.\frac{2}{\sqrt{a}}+\sqrt{b}.\frac{1}{2\sqrt{b}}\right)=\left(2+\frac{1}{2}\right)^2=\frac{25}{4}\)

<=> \(\left(\frac{4}{a}+\frac{1}{ab}\right)\ge5\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{\sqrt{a}}{\frac{2}{\sqrt{a}}}=\frac{\sqrt{b}}{\frac{1}{2\sqrt{b}}}\\a+b=\frac{5}{4}\end{cases}}\)<=>\(\hept{\begin{cases}\frac{a}{2}=2b\\a+b=\frac{5}{4}\end{cases}}\)<=>\(\hept{\begin{cases}x=1\\b=\frac{1}{4}\end{cases}}\)

Vậy min bt = 5 <=> x = 1; b = 1/4

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a^2+b^2+c^2\right)}{3}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}+\frac{2\left(a+b+c\right)^2}{9}\)

\(\ge\frac{\left(\frac{9}{a+b+c}\right)^2}{3}+\frac{2\left(a+b+c\right)^2}{9}=\frac{3^2}{3}+\frac{2.9}{9}=5\)

ko dám đâu

tra chưa

TÔI KHÔNG BIẾT HỎI BẠN LƠPS 9 XEM VÌ TÔI LỚP 4

R=\(2\sqrt{3}\)