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}`

Học giỏi vậy bạn? $x^2-8x+24=(x-2).(x-6)$ ?? Well =)))

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

a đề sai hay sao mà vô nghiệm ?

b)Áp dụng BĐT Cauchy-Schwarz ta có:

\(VP^2=\left(\sqrt{2x+1}+\sqrt{17-2x}\right)^2\)

\(\le\left(1+1\right)\left(2x+1+17-2x\right)=36\)

\(\Rightarrow VP^2\le36\Rightarrow VP\le6\)

Lại có: \(VT=x^4-8x^3+17x^2-8x+22\)

\(=\left(x-4\right)^4+8\left(x-4\right)^3+17\left(x-4\right)^2+6\ge6\)

Thấy: \(VT\le VP=6\)\(\Rightarrow VT=VP=6\)

\(\Rightarrow\left(x-4\right)^4+8\left(x-4\right)^3+17\left(x-4\right)^2+6=6\)

Suy ra x=4

ko hiểu chỗ nào ib nhé

lời giải của bạn trên có 1 xíu sai nhé

Là BĐT Bu-nhi-a Cốp-xki chứ ạ ?

c: ĐKXĐ: \(x^3+3x^2+x-1\ge0\)

=>\(x^3+x^2+2x^2+2x-x-1\ge0\)

=>(x+1)\(\left(x^2+2x-1\right)\ge0\)

=>-1-\(\sqrt2\) <=x<=-1 hoặc \(x\ge-1+\sqrt2\)

\(x^2+5x+2=4\cdot\sqrt{x^3+3x^2+x-1}\)

=>\(x^2-x+6x-6=4\cdot\sqrt{x^3+3x^2+x-1}-8\)

=>(x-1)(x+6)=\(4\cdot\left(\sqrt{x^3+3x^2+x-1}-2\right)=4\cdot\frac{x^3+3x^2+x-1-4}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)(x+6)=\(4\cdot\frac{x^3-x^2+4x^2-4x+5x-5}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)(x+6)=4\(\frac{\left(x-1\right)\left(x^2+4x+5\right)}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)\(\left\lbrack\frac{4\left(x^2+4x+5\right)}{\sqrt{x^3+3x^2+x-1}+2}-x-6\right\rbrack=0\)

=>x-1=0

=>x=1(nhận)

c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}-\left(2-x\right)\left(2+x\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}+\left(x-2\right)\left(2+x\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{2}{1+\sqrt{x-1}}+\left(2+x\right)\right)=0\)

=>x-2=0

=>x=2(nhận)

a:ĐKXĐ: 3<=x<=5

Ta có: \(3x^2-17x+24=\sqrt{x-3}+3\sqrt{5-x}\)

=>\(3x^2-12x-5x+20=\sqrt{x-3}-1+3\sqrt{5-x}-3\)

=>\(\left(x-4\right)\left(3x-5\right)=\frac{x-3-1}{\sqrt{x-3}+1}+3\left(\sqrt{5-x}-1\right)\)

=>\(\left(x-4\right)\left(3x-5\right)=\frac{x-4}{\sqrt{x-3}+1}+3\cdot\frac{5-x-1}{\sqrt{5-x}+1}\)

=>\(\left(x-4\right)\left(3x-5-\frac{1}{\sqrt{x-3}+1}+\frac{3}{\sqrt{5-x}+1}\right)=0\)

=>x-4=0

=>x=4(nhận)

b: ĐKXĐ: x>=1

Ta có: \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2+2-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)+\left(x-2\right)\left(x+2\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}+\left(x-2\right)\left(x+2\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+\frac{-2}{1+\sqrt{x-1}}+\left(x+2\right)\right)=0\)

=>x-2=0

=>x=2(nhận)