a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

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

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

\(\Rightarrow a^2b^2-b^4=b-a\)

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

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

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

ĐKXĐ: \(x\ge1\)

\(\left(\sqrt{x-1}-1\right)+\left(\sqrt{x+7}-3\right)+\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt{x-1}+1}+\dfrac{x-2}{\sqrt{x+7}+3}+\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{\sqrt{x-1}+1}+\dfrac{1}{\sqrt{x+7}+3}+x-1\right)=0\)

\(\Leftrightarrow x-2=0\)

Điều kiện xác định : \(1\le x\le7\)

Bất phương trình chuyển thành :

\(x-1+2\sqrt{7-x}-2\sqrt{x-1}-\sqrt{\left(x-1\right)\left(7-x\right)}\le0\)

Đặt \(a=\sqrt{x-1};b=\sqrt{7-x}\) ta có :

\(a^2-2a-ab+2b\le0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le b\\a\le2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1\le7-x\\x-1\le4\end{matrix}\right.\)

Sau đó tìm x

Điều kiện xác định : \(x\ge2\)

Ta có : \(\sqrt{x+8+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{\left(x+7\right)-\sqrt{x+7}-6}=4\)

\(\Leftrightarrow\sqrt{x+7}+\sqrt{\left(x+7\right)-\sqrt{x+7}-6}-3=0\)

Đặt \(t=\sqrt{x+7},t\ge0\) , pt trở thành \(t+\sqrt{t^2-t-6}-3=0\)

\(\Leftrightarrow\left(t-3\right)+\sqrt{\left(t-3\right)\left(t+2\right)}=0\)

\(\Leftrightarrow\sqrt{t-3}\left(\sqrt{t-3}+\sqrt{t+2}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{t-3}=0\\\sqrt{t-3}+\sqrt{t+2}=0\end{array}\right.\)

Vì \(\sqrt{t-3}\ge0,\sqrt{t+2}\ge0\Rightarrow\sqrt{t-3}+\sqrt{t+2}\ge0\) . Dấu "=" không đồng thời xảy ra nên pt vô nghiệm.

Vậy t = 3 => x = 2

pt có nghiệm x = 2

@Akai Haruma , @phynit giải dùm em vs ạ

ĐKXĐ: \(x>0\)

\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)

\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)

\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)

\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)

\(\Rightarrow a-3>0\Rightarrow a>3\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)

\(\Leftrightarrow2x-6\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)