Đặt \(\dfrac{x}{\sqrt{4x-1}}=a\)

Theo đề, ta có phương trình:

a+1/a=2

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

\(\Leftrightarrow\dfrac{a^2+1-2a}{a}=0\)

=>a=1

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4x-1\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=3\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow x\in\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)

đặt 1/2x-y là a

1/x+y là b

hpt ta đc:

3.a-6.b=1

a-b=0

( giải đi pạn)

Lời giải:

ĐK: $x\geq 0$

Đặt $\sqrt{x+1}=a; \sqrt{x}=b$. ĐK $a,b\geq 0$ thì ta có:

$a-b-ab=a^2-2b^2$

$\Leftrightarrow a-b=a^2+ab-2b^2=(a-b)(a+2b)$

$\Leftrightarrow (a-b)(a+2b-1)=0$

$\Leftrightarrow a=b$ hoặc $a+2b=1$

Nếu $a=b\Rightarrow a^2=b^2\Leftrightarrow x+1=x$ (vô lý)

Nếu $a+2b=1$

$\Leftrightarrow \sqrt{x+1}-1+2\sqrt{x}=0$

$\Leftrightarrow \frac{x}{\sqrt{x+1}+1}+2\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}(\frac{\sqrt{x}}{\sqrt{x+1}+1}+2)=0$

Dễ thấy biểu thức trong ngoặc lớn hơn $0$ nên \sqrt{x}=0$

$\Leftrightarrow x=0$

Vậy.......

Bài này giải kiểu thông thường thì ngắn chứ cưỡng ép đặt ẩn phụ thì nó ko hay, rất dài như dưới đây:

ĐKXĐ: \(xy>0\)

\(\left\{{}\begin{matrix}\dfrac{\sqrt{2}x+\sqrt{2}y}{\sqrt{xy}}=3\\x-y+xy=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{2\left(x+y\right)^2}{xy}}=3\\x-y+xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2\left(x+y\right)^2}{xy}=9\\x-y+xy=3\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x-y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left(x+y\right)^2=\left(x-y\right)^2+4xy=u^2+4v\)

Hệ trở thành:

\(\left\{{}\begin{matrix}\dfrac{2\left(u^2+4v\right)}{v}=9\\u+v=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2u^2+8u=9v\\u+v=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2u^2=v\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2u^2=v\\u+2u^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=2u^2\\2u^2+u-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}u=1\Rightarrow v=2\\u=-\dfrac{3}{2}\Rightarrow v=\dfrac{9}{2}\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}u=1\\v=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=1\\xy=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=x-1\\xy=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-1\\x\left(x-1\right)=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=x-1\\x^2-x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=-2\\x=2\Rightarrow y=1\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}u=-\dfrac{3}{2}\\v=\dfrac{9}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=-\dfrac{3}{2}\\xy=\dfrac{9}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=x+\dfrac{3}{2}\\x\left(x+\dfrac{3}{2}\right)=\dfrac{9}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x+\dfrac{3}{2}\\x^2+\dfrac{3}{2}x-\dfrac{9}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\Rightarrow y=3\\x=-3\Rightarrow y=-\dfrac{3}{2}\end{matrix}\right.\)

ĐKXĐ: x>1 hoặc x<-1

\(x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}\)

=>\(x^2+\frac{2x^2}{\sqrt{x^2-1}}+\frac{x^2}{x^2-1}=\frac{1225}{144}\)

=>\(\frac{x^4-x^2+x^2}{x^2-1}+\frac{2x^2}{\sqrt{x^2-1}}=\frac{1225}{144}\)

=>\(\frac{x^4}{x^2-1}+2\cdot\frac{x^2}{\sqrt{x^2-1}}=\frac{1225}{144}\) (1)

Đặt \(a=\frac{x^2}{\sqrt{x^2-1}}\)

(1) sẽ trở thành: \(a^2+2a=\frac{1225}{144}\)

=>\(144\left(a^2+2a\right)=1225\)

=>\(144a^2+288a-1225=0\)

=>\(144a^2-300a+588a-1225=0\)

=>(12a-25)(12a+49)=0

=>a=25/12(nhận) hoặc a=-49/12(loại)

=>\(\frac{x^2}{\sqrt{x^2-1}}=\frac{25}{12}\)

=>\(25\sqrt{x^2-1}=12x^2\)

=>\(144x^4=625\left(x^2-1\right)=625x^2-625\)

=>\(144x^4-625x^2+625=0\)

=>\(144x^4-225x^2-400x^2+625=0\)

=>\(\left(9x^2-25\right)\left(16x^2-25\right)=0\)

=>\(\left[\begin{array}{l}x^2=\frac{25}{9}\\ x^2=\frac{25}{16}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\left(nhận\right)\\ x=\frac54\left(nhận\right)\end{array}\right.\)

ĐKXĐ : \(1\le x\le3\)

\(x-\sqrt{x-1}-3=0\)

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

Đặt \(t=\sqrt{x-1},t\ge0\), suy ra pt trên trở thành \(t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\Leftrightarrow\orbr{\begin{cases}t=2\left(\text{nhận}\right)\\t=-1\left(\text{loại}\right)\end{cases}}\)

Với t = 2 suy ra x = 5