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

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 \(x=0\)

\(x^4+2x^3=4x+4\)

\(x^4+2x^3+x^2-x^2-4x-4=0\)

\(x^2\left(x^2+2x+1\right)-\left(x^2+4x+4\right)=0\)

\(\left[x\left(x+1\right)\right]^2-\left(x+2\right)^2=0\)

\(\left(x^2+x-x-2\right)\left(x^2+x+2\right)=0\)

\(\left(x^2-2\right)\left(x^2+x+2\right)=0\)

\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x^2+x+2\right)=0\)

tự làm nốt nhé~

\(b,\frac{1}{x^2}+\sqrt{x+2}=\frac{1}{x}+\sqrt{2x+1}\)(1)

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x+2\ge0\\2x+1\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0\\x\ge\frac{-1}{2}\end{cases}}\)

\(\left(1\right)\Leftrightarrow1+x^2\sqrt{x+2}=x+x^2\sqrt{2x+1}\)

\(\Leftrightarrow\left(1-x\right)+x^2\frac{1-x}{\sqrt{x+2}+\sqrt{2x+1}}=0\)

\(\Leftrightarrow\left(1-x\right)\left(1+\frac{x^2}{\sqrt{x+2}+\sqrt{2x+1}}\right)=0\)(2)

Vì\(\hept{\begin{cases}x\ne0\\x\ge\frac{-1}{2}\end{cases}}\Rightarrow1+\frac{x^2}{\sqrt{x+2}+\sqrt{2x+1}}>0\)

Nên từ (2) => Phương trình đã cho có nghiệm x = 1 (TMĐKXĐ)

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

\(\Leftrightarrow7x+4+2\sqrt{\left(2x+1\right)\left(x-1\right)}=3\sqrt{2\left(x-1\right)\left(x+1\right)}+3\sqrt{2\left(2x+1\right)\left(x+1\right)}\)

 \(\Leftrightarrow\left(7x+4+\sqrt{8x^2-4x-4}\right)^2=\left(\sqrt{18x^2-18}+\sqrt{36^2+54x+18}\right)^2\)

\(\Leftrightarrow\left(7x+4\right)^2+8x^2-4x-4+2\left(7x+4\right)\sqrt{8x^2-4x-4}\)\(=18x^2-18+36x^2+54x+18+2\sqrt{\left(18x^2-18\right)\left(36x^2+54x+18\right)}\)

\(\Leftrightarrow3x^2-2x+12+4\left(7x+4\right)\sqrt{\left(x-1\right)\left(2x+1\right)}=36\left(x+1\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow3x^2-2x+12=4\left(2x+5\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow\left(3x^2-2x+12\right)^2=16\left(2x+5\right)^2\left(x-1\right)\left(2x+1\right)\)

\(\Leftrightarrow119x^4+588x^3+1940x^2-672x-544=0\left(1\right)\)

Ta thấy x>1 => Vế trái (1) \(>119.1^4+588.1^3+1940.1^2-672.1-544=1431>0\)

=> pt vô nghiệm.