Question

[Liên Linh]

Giải phương trình 

\(\frac{9-2x}{\sqrt{4-x}}+\frac{4x+3}{\sqrt{4x+1}}=\frac{15}{2}\)

Answer 1

[hung le]

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Answer 2

[Nyatmax]

\(DK:x\in\left(-\frac{1}{4};4\right)\)

PT\(\Leftrightarrow\frac{1}{4}\sqrt{4-x}+\frac{1}{\sqrt{4-x}}+2\sqrt{4x+1}+\frac{2}{\sqrt{4x+1}}+\frac{7}{4}\sqrt{4-x}-\sqrt{4x+1}=\frac{15}{2}\)

Ta co:

\(\frac{1}{4}\sqrt{4-x}+\frac{1}{\sqrt{4-x}}\ge^{ }1\left(1\right)\)

\(2\sqrt{4x+1}+\frac{2}{\sqrt{4x+1}}\ge4\left(2\right)\)

Dau '=' xay ra khi \(x=0\)

Xet

\(\frac{7}{4}\sqrt{4-x}-\sqrt{4x+1}=\frac{5}{2}\left(3\right)\)

\(\Leftrightarrow\frac{-\frac{7}{4}x}{\sqrt{4-x}+2}-\frac{4x}{\sqrt{4x+1}+1}=0\)

\(\Leftrightarrow x\left(\frac{7}{4\sqrt{4-x}+8}+\frac{4}{\sqrt{4x+1}+1}\right)=0\)

\(\Leftrightarrow x=0\left(n\right)\)

Tuc la \(\left(3\right)\)đúng khi \(x=0\) \(\left(4\right)\)

\(\left(1\right),\left(2\right),\left(4\right)\Rightarrow VT\ge\frac{15}{2}=VP\)

Khi \(x=0\)