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:

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

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

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

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

\(\Leftrightarrow...\)

\(x^2+2x\sqrt{x+\dfrac{1}{x}}=8x-1\left(x\ne0\right)\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge\dfrac{1}{8}\)

Vì \(x\ne0\Rightarrow\) chia 2 vế cho x,ta được:

\(x+2\sqrt{x+\dfrac{1}{x}}=8-\dfrac{1}{x}\Rightarrow x+\dfrac{1}{x}+2\sqrt{x+\dfrac{1}{x}}=8\)

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

pt trở thành \(a^2+2a-8=0\Rightarrow a^2-2a+4a-8=0\)

\(\Rightarrow a\left(a-2\right)+4\left(a-2\right)=0\Rightarrow\left(a-2\right)\left(a+4\right)=0\)

mà \(a>0\Rightarrow a=2\Rightarrow\sqrt{x+\dfrac{1}{x}}=2\Rightarrow x+\dfrac{1}{x}=4\)

\(\Rightarrow\dfrac{x^2-4x+1}{x}=0\Rightarrow x^2-4x+1=0\)

\(\Delta=\left(-4\right)^2-4=12\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x-\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{2-\sqrt{3};2+\sqrt{3}\right\}\)

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

ĐK: `{(2x^2+8x+6>=0),(x^2-1>=0),(2x+2>=0):} <=> {(x=-1),(x>=1):}`

`\sqrt(2x^2+8x+6)+\sqrt(x^2-1)=2x+2`

`<=>(2x^2+8x+6)+(x^2-1)+2\sqrt((2x^2+8x+6)(x^2-1))=(2x+2)^2`

`<=>2(x+3)(x+1)+(x-1)(x+2)+2\sqrt((x+1)^2 (x+3)(x-1))=4(x+1)^2`

`<=> (x+1)[2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0`

`<=> [(x=-1\ (TM)),([2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0\ (1)):}`

(1) `<=> x-1=2\sqrt((x+3)(x-1))`

`<=>x^2-2x+1=4(x+3)(x-1)`

`<=>x=1\ `(TM)

Vậy `S={\pm 1}`.

\(ĐK:x\le-3;x\ge-1\)

\(PT\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\\ \Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\\ \Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{25}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)

Vậy \(S=\left\{-1;1\right\}\)

c.

ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)

\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)

- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:

\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)

Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm

- Với \(x\le-5\) pt tương đương:

\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)

Do \(3-x>0\) pt trở thành:

\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)

\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)

\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))

\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)

\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)

a.

Kiểm tra lại đề, pt này không giải được

b.

ĐKXĐ: \(x\ge0\)

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

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

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

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

\(\Leftrightarrow...\)

@Akai Haruma , @phynit giải dùm em vs ạ

Kiểm tra lại đề câu a, \(...+24\) thì pt vô nghiệm, phải là \(...-24\) mới có lý

b/ \(x^2-\left(y+1\right)x+y^2-y-2=0\) (1)

\(\Delta=\left(y+1\right)^2-4\left(y^2-y-2\right)\ge0\)

\(\Leftrightarrow-3y^2+6y+9\ge0\)

\(\Leftrightarrow-1\le y\le3\Rightarrow y=\left\{-1;0;1;2;3\right\}\)

Thay lần lượt vào pt ban đầu để tìm x nguyên

ĐKXĐ: ...

\(\Leftrightarrow x^2+\left(x^2+8x\right)+\left(14-2\sqrt{x^2+8x}\right)x-14\sqrt{x^2+8x}+24=0\)

Đặt \(\sqrt{x^2+8x}=a\ge0\) pt trở thành:

\(x^2+a^2+\left(14-2x\right)x-14a+24=0\)

\(\Leftrightarrow x^2-2ax+a^2+14\left(x-a\right)+24=0\)

\(\Leftrightarrow\left(x-a\right)^2+14\left(x-a\right)+24=0\)

\(\Leftrightarrow\left(x-a+2\right)\left(x-a+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=x+2\\a=x+12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+8x}=x+2\left(x\ge-2\right)\\\sqrt{x^2+8x}=x+12\left(x\ge-12\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+8x=x^2+4x+4\\x^2+8x=x^2+24x+144\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\)