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:

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

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

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

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

\(\Leftrightarrow...\)

@Akai Haruma , @phynit giải dùm em vs ạ

thì v nè

Lời giải:
ĐK: $x\geq -1$

Đặt $\sqrt[3]{9-\sqrt{x+1}}=a; \sqrt[3]{7+\sqrt{x+1}}=b$. Ta có hệ sau đây:

\(\left\{\begin{matrix} a+b=4\\ a^3+b^3=16\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=4\\ (a+b)^3-3ab(a+b)=16\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} a+b=4\\ 64-12ab=16\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=4\\ ab=4\end{matrix}\right.\)

Theo định lý Vi-et đảo, $a,b$ là nghiệm của PT:

$X^2-4X+4=0$

$\Rightarrow a=b=2$

$\Leftrightarrow \sqrt[3]{9-\sqrt{x+1}}=\sqrt[3]{7+\sqrt{x+1}}=2$

$\Rightarrow \sqrt{x+1}=1$

$\Rightarrow x=0$ (thỏa)

Vậy..........

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

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\x\ge-3\\x\ge-4\\x\ge-7\end{matrix}\right.\Leftrightarrow}x\ge-2\)

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

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

\(\Leftrightarrow-2\sqrt{\left(x+2\right)\left(x+3\right)}+2\sqrt{\left(x+4\right)\left(x+7\right)}=6\)

\(\Leftrightarrow2\left[\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}\right]=6\)

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

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}+\left(x+2\right)\left(x+3\right)=9\)

\(\Leftrightarrow-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=-2x^2-16x-8\)

\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=x^2+8x+4\)

Có lẽ làm sai ở đâu đó, mk lười :V

ĐKXĐ: \(x\ge-2\)

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

\(\Leftrightarrow2x+9+2\sqrt{x^2+9x+14}=2x+7+2\sqrt{x^2+7x+12}=0\)

\(\Leftrightarrow\sqrt{x^2+9x+14}+1=\sqrt{x^2+7x+12}\)

\(\Leftrightarrow x^2+9x+15+2\sqrt{x^2+9x+14}=x^2+7x+12\)

\(\Leftrightarrow2\sqrt{x^2+9x+14}=-2x-3\) (\(x\le-\frac{3}{2}\))

\(\Leftrightarrow4\left(x^2+9x+14\right)=4x^2+12x+9\)

\(\Leftrightarrow24x=-47\)

\(\Leftrightarrow x=-\frac{47}{24}\)