a/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{1-x}=a\\\sqrt[4]{x}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}0\le a;b\le1\\a+b=1\\a^4+b^4=1\end{matrix}\right.\)
Do \(0\le a;b\le1\Rightarrow\left\{{}\begin{matrix}a^4\le a\\b^4\le b\end{matrix}\right.\) \(\Rightarrow a^4+b^4\le a+b=1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}a+b=1\\a^4=a\\b^4=b\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(1;0\right);\left(0;1\right)\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[4]{x}=1\\\sqrt[4]{x}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
b/ Đặt \(4x^2-4x+5=a>0\) ta được:
\(\sqrt{a}+\sqrt{3a+4}=6\)
\(\Leftrightarrow4a+4+2\sqrt{3a^2+4a}=36\)
\(\Leftrightarrow\sqrt{3a^2+4a}=16-2a\) (\(a\le8\))
\(\Leftrightarrow3a^2+4a=4a^2-64a+256\)
\(\Leftrightarrow a^2-68a+256=0\Rightarrow\left[{}\begin{matrix}a=4\\a=64\left(l\right)\end{matrix}\right.\)
\(\Rightarrow4x^2-4x+5=4\Leftrightarrow\left(2x-1\right)^2=0\)
b)Ta có:
\(\sqrt{4x^2-4x+5}+\sqrt{12x^2-12x+19}=6\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2+2^2}+\sqrt{3\left(2x-1\right)^2+4^2}=6\)
Vì \(\sqrt{\left(2x-1\right)^2+2^2}\ge2\) và \(\sqrt{3\left(2x-1\right)^2+4^2}\ge4\)
nên \(\sqrt{\left(2x-1\right)^2+2^2}+\sqrt{3\left(2x-1\right)^2+4^2}\ge6\)
Vậy PT \(\left\{{}\begin{matrix}\sqrt{\left(2x-1\right)^2+2^2}=2\\\sqrt{3\left(2x-1\right)^2+4^2}=4\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{1}{2}\)
