Sorry nha nhưng em mới học lớp 7 thôi à ~~

Nhìn không đủ chán rồi không dám động vào

Viết đề kiểu gì v @@

.

a: ĐKXĐ: x>=1

Đặt \(a=\sqrt{5x-1};b=\sqrt{x-1}\)

\(a^2-b^2=\left(5x-1\right)-\left(x-1\right)=4x\)

\(\frac{a^2+b^2}{2}=\frac{5x-1+x-1}{2}=\frac{6x-2}{2}=3x-1\)

\(ab=\sqrt{\left(5x-1\right)\left(x-1\right)}=\sqrt{5x^2-6x+1}\)

Thay vào phương trình ban đầu, ta được: \(\left(a+b\right)\left(\frac{a^2+b^2}{2}-ab\right)=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a+b\right)\left(a-b\right)^2}{2}=\left(a-b\right)\left(a+b\right)\)

=>\(\frac{\left(a-b\right)^2}{2}=\left(a-b\right)\)

=>\(\left(a-b\right)^2=2\left(a-b\right)\)

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

TH1: a-b=0

=>a=b

=>5x-1=x-1

=>x=0(loại)

TH2: a-b-2=0

=>\(\sqrt{5x-1}-\sqrt{x-1}-2=0\)

=>\(\sqrt{5x-1}=\sqrt{x-1}+2\)

=>\(5x-1=x-1+4+2\cdot2\cdot\sqrt{x-1}\)

=>\(4\sqrt{x-1}+x+3=5x-1\)

=>\(4\sqrt{x-1}=5x-1-x-3=4x-4\)

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=2\left(nhận\right)\end{array}\right.\)

\(\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{cases}}\)

\(ĐK:x,y>0\)

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

Vì x, y > 0 nên \(x+2y+\frac{1}{y\sqrt{x}}>0\)suy ra x - y = 0 hay x = y

Thay x = y vào (2), ta được: \(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\)

\(\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}\)\(\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}\)

\(\Leftrightarrow\sqrt{x+3}.\sqrt{x}-\sqrt{x+3}-\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x+3}-1\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\\sqrt{x}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=1\left(tmđk\right)\end{cases}}\Rightarrow x=y=1\)

Vậy hệ có một nghiệm duy nhất \(\left(x;y\right)=\left(1;1\right)\)

\(\hept{\begin{cases}\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{cases}}\)

ĐK: \(\hept{\begin{cases}x>0\\y>0\end{cases}}\)và \(\hept{\begin{cases}x+3\ge0\\x^2+3x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\y>0\end{cases}}}\)

\(\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x+y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0\Leftrightarrow x=y\)do \(x+2y+\frac{1}{y\sqrt{x}}>0\forall x,y>0\)

Thay y=x vào pt (2) ta được

\(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}\)

\(\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}\Leftrightarrow\sqrt{x+3}\cdot\sqrt{x}-\sqrt{x+3}-\sqrt{x+1}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\\sqrt{x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\left(loai\right)\\x=1\left(tm\right)\end{cases}\Rightarrow}x=y=1}\)

Vậy hệ có nghiệm duy nhất (x;y)=(1;1)

Đk: \(\hept{\begin{cases}1-x\ge0\\x^2-3x+2\ge0\\\frac{x-1}{x-2}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\le1\vee x\ge2\\x\le1\vee x\ge2\end{cases}\Leftrightarrow x\le1.}\)

Khi đó \(x-1\le0\Rightarrow x-2\le-1< 0\)

Đặt \(\hept{\begin{cases}\sqrt{1-x}=a\\\sqrt{2-x}=b\end{cases}\Rightarrow\hept{\begin{cases}b^2-a^2=1\\a+ab+\frac{\left(-b^2\right)a}{b}=3\left(1\right)\end{cases}}}\)

Từ (1) \(a+ab-ab=3\Rightarrow a=3\)

\(\Rightarrow\sqrt{1-x}=3\Rightarrow1-x=9\Rightarrow x=-8.\)