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)

https://diendantoanhoc.net/topic/163051-x-fracxsqrtx2-1-frac3512/

\(a,ĐK:x,y\ne2\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\y-2=b\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{3}{a}+\dfrac{2}{b}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{a}+\dfrac{9}{b}=15\\\dfrac{6}{a}+\dfrac{4}{b}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{5}{b}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+3=5\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow x=y=3\left(tm\right)\)

\(b,ĐK:x\ge3;y\ge1\)

Sửa: \(\sqrt{x-3}-\sqrt{y-1}=4\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\ge0\\b=\sqrt{y-1}\ge0\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a-2b=2\\a-b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\-b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3=36\\y-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=39\\y=5\end{matrix}\right.\)

Đặt x+y=a; x-2y=b

=>6/a-3/b=3 và 1/a+7/b=2

=>a=5/3 và b=5

=>x+y=5/3 và x-2y=5

=>x=25/9; y=-10/9

1.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

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

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

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

a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)