\(\sqrt{x-2}+\sqrt{6-x}\text{=}\sqrt{x^2-8x+24}\)

\(ĐKXĐ:2\le x\le6\)

Xét VP của pt ta thấy : \(\sqrt{x^2-8x+24}\text{=}\sqrt{x^2-8x+16+8}\)

\(\text{=}\sqrt{\left(x-4\right)^2+8}\)

\(\Rightarrow VP\ge\sqrt{8}\)

Xét VT của pt ta có :

\(VT^2\text{=}x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

\(VT^2\text{=}4+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Áp dụng BĐT cô si cho 2 số không âm ta có :

\(2\sqrt{\left(x-2\right)\left(6-x\right)}\le\left(\sqrt{x-2}\right)^2+\left(\sqrt{6-x}\right)^2\)

\(\text{=}x-2+6-x\text{=}4\)

\(\Rightarrow VT^2\le8\)

\(\Rightarrow VT\le\sqrt{8}\)

Để \(VT\text{=}VP\) \(\Leftrightarrow\left\{{}\begin{matrix}x-4\text{=}0\\\sqrt{x-2}\text{=}\sqrt{6-x}\end{matrix}\right.\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy...........

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}`

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

ĐKXĐ: \(2\le x\le6\)

\(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\\ \Leftrightarrow\left(\sqrt{x-2}+\sqrt{6-x}\right)^2=\left(\sqrt{x^2-8x+24}\right)^2\\ \Leftrightarrow x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\\ \Leftrightarrow4+2\sqrt{-x^2+8x-12}=x^2-8x+24\\ \Leftrightarrow-x^2+8x-20+2\sqrt{-x^2+8x-12}=0\left(1\right)\)

Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\), ta có:

\(\left(1\right)\Leftrightarrow a^2+2a-8=0\Leftrightarrow\left[{}\begin{matrix}a=2\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)

Ta có:

\(\sqrt{-x^2+8x-12}=2\Leftrightarrow-x^2+8x-12=4\\ \Leftrightarrow-x^2+8x-16=0\\ \Leftrightarrow x^2-8x+16=0\\ \Leftrightarrow\left(x-4\right)^2=0\\ \Leftrightarrow x=4\left(tm\right)\)

Vậy....

P.s: Có gì sai mong mọi người góp ý!

#Lemon

ĐK:....

\(pt\Leftrightarrow x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)

\(\Leftrightarrow4+2\sqrt{-x^2+8x-12}=x^2-8x+24\)

\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)

Đặt \(x^2-8x=a\)

\(pt\Leftrightarrow2\sqrt{-a-12}=a+20\)

\(\Leftrightarrow4\left(-a-12\right)=\left(a+20\right)^2\)

\(\Leftrightarrow a^2+40a+400+4a+48=0\)

\(\Leftrightarrow a^2+44a+448=0\)

\(\Leftrightarrow\left(a+16\right)\left(a+28\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-16\\a=-28\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+16=0\\x^2-8x+28=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\\left(x-4\right)^2+12=0\end{matrix}\right.\)

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

Vậy phương trình có nghiệm duy nhất \(x=4\)

Áp dụng bđt Bunhia,ta có VT^2<=2(x-2+6-x)=8

suy ra VT<=\(2\sqrt{2}\)

Dấu "=" xảy ra khi \(\sqrt{x-2}=\sqrt{6-x}\) <=> x-2=6-x <=>x=4

Mặc khác \(\sqrt{x^2-8x+24}=\sqrt{\left(x-4\right)^2+8}>=2\sqrt{2}\)

Dấu "=" xảy ra khi \(\left(x-4\right)^2\)=0 <=> x=4

Vậy pt đã cho có 1 nghiệm duy nhất là x=4

ĐKXĐ: \(x\ge0\)

\(\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)