đề bài đúng không z? theo tôi đề là \(\sqrt{x+2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)?!

ĐKXĐ:...

Áp dụng BĐT AM-GM:

\(\left(\sqrt{x+2}+\sqrt{6-x}\right)^2\le2\left(x+2+6-x\right)=16\)

\(\Leftrightarrow\sqrt{x+2}+\sqrt{6-x}\le4\)

Lại có \(x^2-8x+24=\left(x-4\right)^2+8\ge8\forall x\)

Vậy pt vô nghiệm.

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:

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

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

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

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

\(\Leftrightarrow...\)

Lag tí -.-'

`ĐK:2<=x<=6`

BP 2 vế ta có:

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

`<=>4+2\sqrt{(x-2)(6-x)}=x^2-8x+24`

`<=>2\sqrt{(x-2)(6-x)}=x^2-8x+20`

`<=>2sqrt{-x^2+8x-12}=x^2-8x+20`

`<=>-x^2+8x-20+2sqrt{-x^2+8x-12}=0`

`<=>-x^2+8x-12+2sqrt{-x^2+8x-12}-8=0`

Đặt `sqrt{-x^2+8x-12}=a(a>=0)`

`pt<=>a^2+2a-8=0`

`<=>a=2(tm),a=-4(l)`

`<=>-x^2+8x-12=4`

`<=>x^2-8x+16=0`

`<=>(x-4)^2=0<=>x=4(tmđk)`

Vậy `S={4}`

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

\(ĐKXĐ:2\le x\le6\)

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...........

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: \(x\ge0\)

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

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

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

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

\(\Leftrightarrow...\)

=>-(x+3)^2*(x-4)(x+12)=x^2-48x+576

=>-(x^2+6x+9)(x^2+8x-48)=x^2-48x+576

=>-x^4-14x^3-9x^2+216x+432=x^2-48x+576

=>x^4+14x^3+10x^2-264x+144=0

=>(x^2+4x-24)(x^2+10x-6)=0

=>\(x\in\left\{-5+\sqrt{31};-5-\sqrt{31};-2+2\sqrt{7};-2-2\sqrt{7}\right\}\)

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 \(S=\left\{-1;1\right\}\)