a/ ĐKXĐ: \(x>\frac{1}{2}\)
\(\Leftrightarrow\frac{3x^2-1}{\sqrt{2x-1}}-\sqrt{2x-1}=mx\)
\(\Leftrightarrow\frac{3x^2-2x}{\sqrt{2x-1}}=mx\Leftrightarrow\frac{3x-2}{\sqrt{2x-1}}=m\)
Đặt \(\sqrt{2x-1}=a>0\Rightarrow x=\frac{a^2+1}{2}\Rightarrow\frac{3a^2-1}{2a}=m\)
Xét hàm \(f\left(a\right)=\frac{3a^2-1}{2a}\) với \(a>0\)
\(f'\left(a\right)=\frac{12a^2-2\left(3a^2-1\right)}{4a^2}=\frac{6a^2+2}{4a^2}>0\)
\(\Rightarrow f\left(a\right)\) đồng biến
Mặt khác \(\lim\limits_{a\rightarrow0^+}\frac{3a^2-1}{2a}=-\infty\); \(\lim\limits_{a\rightarrow+\infty}\frac{3a^2-1}{2a}=+\infty\)
\(\Rightarrow\) Phương trình đã cho luôn có nghiệm với mọi m
b/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\sqrt[4]{\left(x-1\right)^2}+4m\sqrt[4]{\left(x-1\right)\left(x-2\right)}+\left(m+3\right)\sqrt[4]{\left(x-2\right)^2}=0\)
Nhận thấy \(x=2\) không phải là nghiệm, chia 2 vế cho \(\sqrt[4]{\left(x-2\right)^2}\) ta được:
\(\sqrt[4]{\left(\frac{x-1}{x-2}\right)^2}+4m\sqrt[4]{\frac{x-1}{x-2}}+m+3=0\)
Đặt \(\sqrt[4]{\frac{x-1}{x-2}}=a\) pt trở thành: \(a^2+4m.a+m+3=0\) (1)
Xét \(f\left(x\right)=\frac{x-1}{x-2}\) khi \(x>0\)
\(f'\left(x\right)=\frac{-1}{\left(x-2\right)^2}< 0\Rightarrow f\left(x\right)\) nghịch biến
\(\lim\limits_{x\rightarrow2^+}\frac{x-1}{x-2}=+\infty\) ; \(\lim\limits_{x\rightarrow+\infty}\frac{x-1}{x-2}=1\) \(\Rightarrow f\left(x\right)>1\Rightarrow a>1\)
\(\left(1\right)\Leftrightarrow m\left(4a+1\right)=-a^2-3\Leftrightarrow m=\frac{-a^2-3}{4a+1}\)
Xét \(f\left(a\right)=\frac{-a^2-3}{4a+1}\) với \(a>1\)
\(f'\left(a\right)=\frac{-2a\left(4a+1\right)-4\left(-a^2-3\right)}{\left(4a+1\right)^2}=\frac{-4a^2-2a+12}{\left(4a+1\right)^2}=0\Rightarrow a=\frac{3}{2}\)
\(f\left(1\right)=-\frac{4}{5};f\left(\frac{3}{2}\right)=-\frac{3}{4};\) \(\lim\limits_{a\rightarrow+\infty}\frac{-a^2-3}{4a+1}=-\infty\)
\(\Rightarrow f\left(a\right)\le-\frac{3}{4}\Rightarrow m\le-\frac{3}{4}\)
c/ ĐKXĐ: \(-5\le x\le4\)
Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
\(\Rightarrow\sqrt{4-x}+\sqrt{x+5}\ge\sqrt{4-x+x+5}=3\)
Áp dụng BĐT Bunhiacôpxki:
\(\sqrt{4-x}+\sqrt{x+5}\le\sqrt{\left(1+1\right)\left(4-x+x+5\right)}=3\sqrt{2}\)
\(\Rightarrow3\le m\le3\sqrt{2}\)
d/
\(\Leftrightarrow m\left(\sqrt{2x^2+9}-1\right)< x\)
Do \(\sqrt{2x^2+9}\ge\sqrt{9}=3\Rightarrow\sqrt{2x^2+9}-1>0\) nên BPT tương đương:
\(\Leftrightarrow m< \frac{x}{\sqrt{2x^2+9}-1}\)
Đặt \(f\left(x\right)=\frac{x}{\sqrt{2x^2+9}-1}\)
Để BPT đã cho có nghiệm thì \(m< \max\limits_{x\in R}f\left(x\right)\)
\(f'\left(x\right)=\frac{\sqrt{2x^2+9}-1-\frac{2x^2}{\sqrt{2x^2+9}}}{\left(\sqrt{2x^2+9}-1\right)^2}=\frac{9-\sqrt{2x^2+9}}{\left(\sqrt{2x^2+9}-1\right)^2\sqrt{2x^2+9}}\)
\(f'\left(x\right)=0\Rightarrow2x^2+9=81\Rightarrow\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\)
\(f\left(-6\right)=-\frac{3}{4};f\left(6\right)=\frac{3}{4}\) ; \(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=\frac{\sqrt{2}}{2}\); \(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=\frac{-\sqrt{2}}{2}\)
\(\Rightarrow\max\limits_{x\in R}f\left(x\right)=\frac{3}{4}\Rightarrow m< \frac{3}{4}\)