\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)

\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)

\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)

==>Sai đề không mem

(x-2)² = 25 => x- 2 = 5
                  => x = 7

\(x^2-4x+4=25\)

\(\Leftrightarrow\left(x-2\right)^2=25\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

=0,7320721383

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

<=>  \(\frac{ab+bc+ca}{abc}=0\)

<=>  \(ab+bc+ca=0\)

=>  \(ab+bc=-ca\)

<=>  \(\left(ab+bc\right)^3=-ca^3\)

Ta co:   \(a^3b^3+b^3c^3+c^3a^3=a^3b^3+b^3c^3-\left(ab+bc\right)^3=a^3b^3+b^3c^3-ab^3-bc^3-3ab.bc\left(ab+bc\right)\)

\(=-3ab.bc\left(ab+bc\right)=-3ab.bc.\left(-ca\right)=3a^2b^2c^2\)

\(M=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{b^3c^3+c^3a^3+a^3b^3}{abc}=\frac{3a^2b^2c^2}{abc}=3abc\)

\(5\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)

\(B_{min}\Leftrightarrow2-5\sqrt{x}ĐạtGTLNvà:\sqrt{x}+3đạtGTNN\)

Dấu "=" xảy ra <=> x=0

Vậy B_min=2/3

Ghi nốt đề đi