Điều kiện xác định x#1; y#3.Đặt: \(\hept{\begin{cases}\frac{1}{x-1}=a\\\frac{1}{y-3}=b\end{cases}}\Rightarrow\hept{\begin{cases}5a+b=10\\a-3b=18\end{cases}}\Rightarrow\hept{\begin{cases}15a+3b=30\\a-3b=18\end{cases}}\)

Cộng theo vế: \(15a+3b+a-3b=48\Rightarrow16a=48\Rightarrow a=3\Rightarrow b=-5\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{x-1}=3\Rightarrow x=\frac{4}{3}\\\frac{1}{y-3}=-5\Rightarrow y=-\frac{14}{5}\end{cases}}\)

\(\hept{\begin{cases}\frac{5}{x-1}+\frac{1}{y-3}=10\\\frac{1}{x-1}-\frac{3}{y-3}=18\end{cases}}\)

Đặt: \(\frac{1}{x-1}=a\left(a>0\right);\frac{1}{y-3}=b\left(b>0\right)\)

Khi đó hpt có dạng:

\(\hept{\begin{cases}5a+b=10\\a-3b=18\end{cases}\Rightarrow\hept{\begin{cases}a=3\\b=-5\end{cases}}\left(Tm\right)}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{x-1}=3\\\frac{1}{y-3}=-5\end{cases}}\Rightarrow\hept{\begin{cases}3\left(x-1\right)=1\\-5\left(y-3\right)=1\end{cases}}\Rightarrow\hept{\begin{cases}3x-3=1\\-5y+15=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{14}{5}\end{cases}}\)

\(\hept{\begin{cases}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{cases}}\)

Đặt \(\sqrt{x-1}=a\left(a\ge0\right)\)

\(\sqrt{y+2}=b\left(b\ge0\right)\)

Khi đó hpt có dạng:

\(\hept{\begin{cases}a-3b=2\\2a+5b=15\end{cases}\Rightarrow\hept{\begin{cases}2a-6b=4\\2a+5b=15\end{cases}}\Rightarrow\hept{\begin{cases}-11b=-11\\2a+5b=15\end{cases}}\Rightarrow\hept{\begin{cases}b=1\\2a+5.1=15\end{cases}}\Rightarrow\hept{\begin{cases}b=1\\a=5\end{cases}\left(TM\right)}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x-1}=5\\\sqrt{y+2}=1\end{cases}\Rightarrow\hept{\begin{cases}x-1=25\\y+2=1\end{cases}\Rightarrow}\hept{\begin{cases}x=26\\y=-1\end{cases}}}\)

\(\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x}\)\(\text{ ĐKXĐ : }0< x\le1\)

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

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

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

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

\(\Leftrightarrow1-2x=0\) \(\left(\text{Vì biểu thức trong ngoặc vuông luôn lớn hơn 0 với mọi x}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)

\(\text{Vậy pt có nghiệm }x=\frac{1}{2}\)

#)Giải :

a) Câu trc của bn mk có giải rùi, thắc mắc vô Thống kê hđ của mk xem lại nhé !

b) Để \(P>0\Rightarrow\frac{x-1}{\sqrt{x}}>0\Rightarrow x-1>0\left(\sqrt{x}>0\right)\Rightarrow x>1\)

c) Bó tay @@

\(a,P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(x-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x-1}{\sqrt{x}}\)

Vậy với \(x>0;x\ne1\)thì \(P=\frac{x-1}{\sqrt{x}}\)

\(b,\)Để \(P>0\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\Leftrightarrow x-1>0\Leftrightarrow x>1\left(\sqrt{x}>0\right)\)

#)Giải :

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(P=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\div\frac{1}{\sqrt{x}-1}=\frac{x-1}{\sqrt{x}}\)

\(7\sqrt{2x}-2\sqrt{2x}-4=3\sqrt{2x}\)

\(7\sqrt{2x}-2\sqrt{2x}-3\sqrt{2x}=4\)

\(2\sqrt{2x}=4\)

\(\sqrt{2x}=\frac{4}{2}=2=\sqrt{4}\)

\(\rightarrow2x=4\rightarrow x=2\)

\(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

\(\sqrt{\left(x-3\right)^2}=\sqrt{1+2.1\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

\(\left|x-3\right|=1+\sqrt{3}\)

Chia 2 TH

Với x lớn hơn hoặc bằng 3 => \(x=4+\sqrt{3}\)

Với x bé hơn 3 => \(x=2+\sqrt{3}\)