ĐKXĐ: \(\left[{}\begin{matrix}-1\le x< 0\\x\ge\sqrt{\dfrac{5}{2}}\end{matrix}\right.\)

\(x-\dfrac{4}{x}+\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}=0\)

\(\Leftrightarrow x-\dfrac{4}{x}+\dfrac{x-\dfrac{4}{x}}{\sqrt{2x-\dfrac{5}{x}}+\sqrt{x-\dfrac{1}{x}}}=0\)

\(\Leftrightarrow\left(x-\dfrac{4}{x}\right)\left(1+\dfrac{1}{\sqrt{2x-\dfrac{5}{x}}+\sqrt{x-\dfrac{1}{x}}}\right)=0\)

\(\Leftrightarrow x-\dfrac{4}{x}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\left(loại\right)\end{matrix}\right.\)

Đk: \(\left[{}\begin{matrix}-1\le x< 0\\\dfrac{\sqrt{10}}{2}\le x\le2\end{matrix}\right.\)

Phương trình đã cho trở thành:

\(\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}+x-\dfrac{4}{x}=0\left(\cdot\right)\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{2x-\dfrac{5}{x}}\left(a>0\right)\\b=\sqrt{x-\dfrac{1}{x}}\left(b>0\right)\end{matrix}\right.\)

\(\left(\cdot\right)\Rightarrow a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=-1\left(voli\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x-\dfrac{5}{x}}=\sqrt{x-\dfrac{1}{x}}\)

\(\Rightarrow2x-\dfrac{5}{x}=x-\dfrac{1}{x}\)

\(\Leftrightarrow x-\dfrac{4}{x}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm duy nhất \(x=2\)

ĐKXĐ: x>0

\(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)

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

Đặt \(a=\sqrt{x}+\frac{1}{2\sqrt{x}}\)

=>\(a^2=x+\frac{1}{4x}+2\cdot\sqrt{x}\cdot\frac{1}{2\sqrt{x}}=x+\frac{1}{4x}+1\)

=>\(2a^2=2x+\frac{1}{2x}+2\)

=>\(2a^2-2=2x+\frac{1}{2x}\)

(1) sẽ tương đương với: 5a=\(2a^2-2+4=2a^2+2\)

=>\(2a^2-5a+2=0\)

=>(2a-1)(a-2)=0

=>a=1/2 hoặc a=2

TH1: a=1/2

=>\(\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac12\)

=>\(\frac{2x+1}{2\sqrt{x}}=\frac{\sqrt{x}}{2\sqrt{x}}\)

=>\(2x+1=\sqrt{x}\)

=>\(2x-\sqrt{x}+1=0\)

=>\(x-\frac12\cdot\sqrt{x}+\frac12=0\)

=>\(x-2\cdot\sqrt{x}\cdot\frac14+\frac{1}{16}+\frac{7}{16}=0\)

=>\(\left(\sqrt{x}-\frac14\right)^2+\frac{7}{16}=0\)

=>Loại

TH2: a=2

=>\(\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)

=>\(\frac{2x+1}{2\sqrt{x}}=2\)

=>\(2x-4\sqrt{x}+1=0\)

=>\(x-2\sqrt{x}+\frac12=0\)

=>\(x-2\sqrt{x}+1-\frac12=0\)

=>\(\left(\sqrt{x}-1\right)^2=\frac12\)

=>\(\left[\begin{array}{l}\sqrt{x}-1=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\\ \sqrt{x}-1=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=1+\frac{\sqrt2}{2}=\frac{2+\sqrt2}{2}\\ \sqrt{x}=1-\frac{\sqrt2}{2}=\frac{2-\sqrt2}{2}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{\left(2+\sqrt2\right)^2}{2^2}=\frac{6+4\sqrt2}{4}=\frac{3+2\sqrt2}{2}\\ x=\frac{\left(2-\sqrt2\right)^2}{2^2}=\frac{6-4\sqrt2}{4}=\frac{3-2\sqrt2}{2}\end{array}\right.\)

b.

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

\(\Leftrightarrow x^4+2x^2-4+x\sqrt{2x^2+4x}=0\)

Đặt \(x\sqrt{2x^2+4x}=t\Rightarrow t^2=x^2\left(2x^2+4x\right)=2\left(x^4+2x^2\right)\)

Pt trở thành:

\(\dfrac{t^2}{2}-4+t=0\)

\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4x}=2\left(x>0\right)\\x\sqrt{2x^2+4x}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-2=0\left(x>0\right)\\x^4+2x^2-8=0\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)

a.

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\dfrac{9}{x^2}+2+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

\(\Leftrightarrow\dfrac{2x^2+9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

Đặt \(\dfrac{x}{\sqrt{2x^2+9}}=t\Rightarrow\dfrac{2x^2+9}{x^2}=\dfrac{1}{t^2}\)

Pt trở thành:

\(\dfrac{1}{t^2}+2t=3\)

\(\Rightarrow2t^3-3t^2+1=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(2t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{2x^2+9}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{2x^2+9}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vô-nghiệm\right)\\4x^2=2x^2+9\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{3\sqrt{2}}{2}\)

Kiểm tra lại vế trái đề bài câu b

a, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...