Ghi thiếu đề bài nên tl lại 

`sqrt{x-2}+sqrt{6-x}=x^2-8x+16+2sqrt2`

Áp dụng BĐT bunhia ta có:

`sqrt{x-2}+sqrt{6-x}<=sqrt{(1+1)(x-2+6-x)}=2sqrt2`

`=>VT<=2sqrt2(1)`

Mặt khác:

`VP=x^2-8x+16+2sqrt2`

`=(x-4)^2+2sqrt2>=2sqrt2`

`=>VP>=2sqrt2(2)`

`(1)(2)=>VT=VP=2sqrt2`

`<=>x=4`

Vậy `S={4}`

`sqrt{x-2}+sqrt{6-x}=x^2-8x+2sqrt2`

Áp dụng BĐT bunhia ta có:

`sqrt{x-2}+sqrt{6-x}<=sqrt{(1+1)(x-2+6-x)}=2sqrt2`

`=>VT<=2sqrt2(1)`

Mặt khác:

`VP=x^2-8x+16+2sqrt2`

`=(x-4)^2+2sqrt2>=2sqrt2`

`=>VP>=2sqrt2(2)`

`(1)(2)=>VT=VP=2sqrt2`

`<=>x=4`

Vậy `S={4}`

Đề đúng chưa v

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1

\(a,ĐK:-9\le x\le16\\ PT\Leftrightarrow\left(\sqrt{16-x}-3\right)+\left(\sqrt{x+9}-4\right)=0\\ \Leftrightarrow\dfrac{7-x}{\sqrt{16-x}+3}+\dfrac{x-7}{\sqrt{x+9}+4}=0\\ \Leftrightarrow\left(x-7\right)\left(\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}=0\end{matrix}\right.\)

Với \(x\ge-9\) thì \(\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}>0\)

Do đó PT có nghiệm duy nhất \(x=7\)

\(b,ĐK:-\sqrt{2}\le x\le\sqrt{2}\\ PT\Leftrightarrow\left(\sqrt{2-x^2}-1\right)+\left(\sqrt{x^2+8}-3\right)=0\\ \Leftrightarrow\dfrac{1-x^2}{\sqrt{2-x^2}+1}+\dfrac{x^2-1}{\sqrt{x^2+8}+3}=0\\ \Leftrightarrow\left(x^2-1\right)\left(\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}=0\end{matrix}\right.\)

Với \(x\ge-\sqrt{2}\) thì \(\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}>0\)

Vậy pt có tập nghiệm \(x=\pm1\)

a) Đk: \(\left\{{}\begin{matrix}16-x\ge0\\x+9\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le16\\x\ge-9\end{matrix}\right.\) \(\Rightarrow x\in\left[-9;16\right]\)

    Pt: \(\Rightarrow\left(\sqrt{16-x}+\sqrt{x+9}\right)^2=7^2\)

         \(\Rightarrow16-x+x+9+2\sqrt{144+7x-x^2}=49\)

         \(\Rightarrow\sqrt{144+7x-x^2}=12\)

         \(\Rightarrow144+7x-x^2=144\)

 Bạn tự tìm x nhé rồi đối chiếu đk ta đc \(x=0\) hoặc \(x=7\)

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

\(\Leftrightarrow\left(\sqrt{x^2+16}-5\right)+\left(4-\sqrt{x^2+7}\right)=3x-9\)

\(\Leftrightarrow\frac{x^2-9}{\sqrt{x^2+16}+5}+\frac{9-x^2}{\sqrt{x^2+7}+4}=3\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{x+3}{\sqrt{x^2+16}+5}-\frac{x+3}{\sqrt{x^2+7}+4}-3\right)=0\)

\(\Leftrightarrow x=3\)

Đk: `1 <=x <=7`.

Đặt `sqrt(7-x) = a, sqrt(x-1) = b`.

Phương trình trở thành: `b^2+1 + 2a = 2b + ab + 1`.

`<=> b^2 + 2a = 2b + ab.`

`<=> b(b-2) = a(b-2)`

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

`<=> a =b` hoặc `b = 2.`

`@ a = b => 7 - x = x - 1`

`<=> 8 = 2x <=> x = 4`.

`@ b = 2 => sqrt(x-1) = 2`

`<=> x - 1 = 4`

`<=> x = 5`.

Vậy `x = 4` hoặc `x = 5`.

\(\text{ĐKXĐ:}1\le x\le7\)

PT đã cho tương đương với:

\(x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{x-1}.\sqrt{7-x}=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{4;5\right\}\)

\(\Leftrightarrow\sqrt{x+4}\left(\sqrt{x-4}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-4=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=13\end{matrix}\right.\)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x=-4\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{\left(x-4\right)\left(x+4\right)}-3\sqrt{x+4}=0\)

\(\Leftrightarrow\sqrt{x+4}.\left(\sqrt{x-4}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}=0\\\sqrt{x-4}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-4=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(tm\right)\\x=13\left(tm\right)\end{matrix}\right.\)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x=-4\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\left(x-4\right)\left(x+4\right)}=3\sqrt{\left(x+4\right)}\\ \Leftrightarrow\left(x-4\right)\left(x+4\right)=9\left(x+4\right)\\ \Leftrightarrow\left(x+4\right)\left(x-13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\left(tm\right)\\x=13\left(tm\right)\end{matrix}\right.\)