Để hàm số trên là hàm số đồng biến khi \(1-3m>0\Leftrightarrow m< \frac{1}{3}\)

Để hàm số trên là hàm số nghịch biến khi \(1-3m< 0\Leftrightarrow m>\frac{1}{3}\)

\(10+\sqrt{x^2+6x+10}\)

\(10+\sqrt{\left(x+3\right)^2+1}\ge10+\sqrt{1}=11\)

dấu "=" xảy ra khi và chỉ khi \(x+3=0< =>x=-3\)

\(< =>MIN=11\)

a, \(A=\frac{1}{\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}}\)

\(=\frac{\sqrt{4+\sqrt{10+2\sqrt{5}}}-\sqrt{4-\sqrt{10+2\sqrt{5}}}}{4+\sqrt{10+2\sqrt{5}}-4+\sqrt{10+2\sqrt{5}}}\)

\(=\frac{\sqrt{4+\sqrt{10+2\sqrt{5}}}-\sqrt{4-\sqrt{10+2\sqrt{5}}}}{2\sqrt{10+2\sqrt{5}}}\)

\(A^2=\frac{4+\sqrt{10+2\sqrt{5}}-2\sqrt{16-10-2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}}{4\left(10+2\sqrt{5}\right)}\)

\(=\frac{8-2\sqrt{6-2\sqrt{5}}}{40+8\sqrt{5}}=\frac{9-2\left(\sqrt{5}-1\right)}{40+2.4\sqrt{5}}\)

\(\Rightarrow A=\sqrt{\frac{11-2\sqrt{5}}{40+8\sqrt{5}}}=\frac{\sqrt{11-2\sqrt{5}}}{2\sqrt{10+2\sqrt{5}}}=\frac{\sqrt{\left(11-2\sqrt{5}\right)\left(10+2\sqrt{5}\right)}}{20+4\sqrt{5}}\)

\(=\frac{\sqrt{110+2\sqrt{5}-20}}{20+4\sqrt{5}}=\frac{\sqrt{90+2\sqrt{5}}}{20+4\sqrt{5}}\)

trục căn thức cho biểu thức mất căn là được 

Sửa 

\(A^2=\frac{1}{4+\sqrt{10+2\sqrt{5}}+2\sqrt{16-10-2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}}\)

\(=\frac{1}{8+2\sqrt{6-2\sqrt{5}}}=\frac{1}{8+2\left(\sqrt{5}-1\right)}=\frac{1}{6+2\sqrt{5}}\)

\(\Rightarrow A=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}\)

e, Đặt \(A=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=3-\sqrt{5}+2\sqrt{9-5}+3+\sqrt{5}=6+2.2=10\)

a, \(A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)ĐK : \(a>0;a\ne1\)

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

\(=\frac{\left(a-1\right)^2}{4a}.\frac{-4\sqrt{a}}{a-1}=\frac{1-a}{\sqrt{a}}\)

b, \(\frac{1-a}{\sqrt{a}}< 0\Rightarrow\hept{\begin{cases}1-a< 0\\\sqrt{a}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>1\\a>0\end{cases}}\Leftrightarrow a>1\)

c, Ta có : \(A=-2\Rightarrow\frac{1-a}{\sqrt{a}}=-2\Rightarrow1-a=-2\sqrt{a}\)

\(\Leftrightarrow a-2\sqrt{a}-1=0\Leftrightarrow\left(\sqrt{a}-1\right)^2-2=0\)

\(\Leftrightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)

\(\Rightarrow\sqrt{a}=1+\sqrt{2}\left(tm\right);1-\sqrt{a}\left(ktm\right)\Leftrightarrow a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\)

\(\frac{2xy^2}{3ab}\sqrt{\frac{9a^3b^4}{8xy^3}}=\frac{2xy^2}{3ab}\frac{3\sqrt{a^2.a}\sqrt{\left(b^2\right)^2}}{2\sqrt{2xy^2.y}}\)

\(=\frac{2xy^2}{3ab}\frac{3a\sqrt{a}b^2}{2y\sqrt{2xy}}=\frac{6xy^2ab^2\sqrt{a}}{6aby\sqrt{2xy}}=\frac{bxy\sqrt{a}}{\sqrt{2xy}}\)

\(=\frac{bxy\sqrt{2axy}}{2xy}=\frac{b\sqrt{2axy}}{2}\)