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ĐKXĐ \(2\le x\le4\).Đặt A=\(\sqrt[4]{\left(x-2\right)\left(4-x\right)}+\sqrt[4]{x-2}+\sqrt[4]{4-x}+6x\sqrt{3x}\)
Do x\(\ge2>0\)nên ADBĐT CAUCHY ta được:
\(\sqrt[4]{1\cdot1\cdot\left(x-2\right)\left(4-x\right)}\le\frac{1+1+x-2+4-x}{4}=1\)
\(\sqrt[4]{x-2}\le\frac{1+1+1+x-2}{4}=\frac{1}{4}\)
\(\sqrt[4]{4-x}\le\frac{1+1+1+4-x}{4}=\frac{7}{4}\)
\(6x\sqrt{3x}=2\sqrt{27x^3}\le x^3+27\)
_Do đó A\(\le1+\frac{1}{4}+\frac{7}{4}+x^3+27=x^3+30\)
Dấu = xảy ra \(\Leftrightarrow x=3\)(thỏa mãn ĐKXĐ)
\(A=\left(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}}\right)\sqrt{x+\sqrt{x^2-32}}\) với \(x\ge4\sqrt{2}\)
Lời giải:
\(A\sqrt{2}=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{2x+\sqrt{(x-4\sqrt{2})(x+4\sqrt{2})}}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})^2}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})\)
\(=(\sqrt{x-4\sqrt{2}})^2-(\sqrt{x+4\sqrt{2}})^2=(x-4\sqrt{2})-(x+4\sqrt{2})=-8\sqrt{2}\)
Lời giải:
\(A\sqrt{2}=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{2x+\sqrt{(x-4\sqrt{2})(x+4\sqrt{2})}}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})\sqrt{(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})^2}\)
\(=(\sqrt{x-4\sqrt{2}}-\sqrt{x+4\sqrt{2}})(\sqrt{x-4\sqrt{2}}+\sqrt{x+4\sqrt{2}})\)
\(=(\sqrt{x-4\sqrt{2}})^2-(\sqrt{x+4\sqrt{2}})^2=(x-4\sqrt{2})-(x+4\sqrt{2})=-8\sqrt{2}\)
Bài 1:
a) \(\sqrt{1-x^2}\)có nghĩa \(\Leftrightarrow\)\(1-x^2\ge0\)
\(\Leftrightarrow\)\(x^2\le1\)
\(\Leftrightarrow\)\(\left|x\right|\le1\)
b) \(\sqrt{\frac{x-2}{x-3}}\)có nghĩa \(\Leftrightarrow\)\(\frac{x-2}{x-3}\ge0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x>3\\x\le2\end{cases}}\)
a/ ĐKXĐ: \(x\ge\frac{3}{4}\)
\(\Leftrightarrow6x+1+2\sqrt{5x^2+5x}=6x+1+2\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\)
\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)
\(\Leftrightarrow3x^2+5x-12=0\Rightarrow\left[{}\begin{matrix}x=-3< \frac{3}{4}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
b/ \(\Leftrightarrow x^2+x+1+2\sqrt{x^2+x+1}-3=0\)
Đặt \(\sqrt{x^2+x+1}=t>0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+x+1}=1\)
\(\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a) ĐKXD:...
\(pt\Leftrightarrow\left(\sqrt{x+2}+\sqrt{x-2}\right)^2=6-2x\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{x-2}=\sqrt{6-2x}\)
Đến đây dễ rồi
ĐKXĐ: \(-4\le x\le1\)
Đặt \(\sqrt{x+4}-\sqrt{1-x}=t\)
\(\Rightarrow t^2=5-2\sqrt{\left(x+4\right)\left(1-x\right)}\Rightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=\frac{5-t^2}{2}\)
Pt trở thành:
\(t\left(1+\frac{5-t^2}{2}\right)=3\Leftrightarrow t\left(7-t^2\right)=6\)
\(\Leftrightarrow t^3-7t+6=0\Leftrightarrow\left(t+3\right)\left(t-1\right)\left(t-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=1\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+4}-\sqrt{1-x}=-3\\\sqrt{x+4}-\sqrt{1-x}=1\\\sqrt{x+4}-\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}+3=\sqrt{1-x}\left(vn\right)\\\sqrt{x+4}=1+\sqrt{1-x}\\\sqrt{x+4}=2+\sqrt{1-x}\end{matrix}\right.\) (1 vô nghiệm do \(VT\ge3;VP\le\sqrt{5}< 3\))
\(\Leftrightarrow\left[{}\begin{matrix}x+4=2-x+2\sqrt{1-x}\\x+4=5-x+4\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{1-x}\left(x\ge-1\right)\\2x-1=4\sqrt{1-x}\left(x\ge\frac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=1-x\\4x^2-4x+1=16-16x\end{matrix}\right.\) \(\Leftrightarrow...\)