pt <=> \(2x^2-20x+54-2\sqrt{x-4}-2\sqrt{6-x}=0\)

<=> \(\left(2x^2-20x+50\right)+\left(x-4-2\sqrt{x-4}+1\right)+\left(6-x-2\sqrt{6-x}+1\right)=0\)

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

<=> x = 5

ĐKXĐ : \(4\le x\le6\)

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

Áp dụng bđt Cauchy ta có : \(2+2\sqrt{\left(6-x\right)\left(x-4\right)}\le2+6-x+x-4=4\)

\(\Rightarrow VP\le2\forall x\)(1)

Xét \(VT=x^2-10x+27=\left(x^2-10x+25\right)+2=\left(x-5\right)^2+2\ge2\forall x\)(2)

Từ (1);(2) \(\Rightarrow VT\ge2\ge VP\)

Dấu "=" xảy ra \(\hept{\begin{cases}6-x=x-4\\\left(x-5\right)^2=0\end{cases}\Rightarrow x=5\left(TMĐKXĐ\right)}\)

Vậy nghiệm pt là x = 5

Áp dụng bđt AM-GM:

\(\sqrt{x-4}\le\dfrac{x-4+1}{2}=\dfrac{x-3}{2}\)

\(\sqrt{6-x}\le\dfrac{6-x+1}{2}=\dfrac{7-x}{2}\)

Cộng theo vế: \(VT\le\dfrac{x-3+7-x}{2}=2\)

Mặt khác: \(VP=x^2-10x+27=\left(x-5\right)^2+2\ge2\)

\(VT=VP\Leftrightarrow x=5\)

ta có đề bài <=> 

\(\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\)

<=> \(\left|x-3\right|+\left|x+5\right|=8\)

<=>\(\left|3-x\right|+\left|x+5\right|=8\)

Áp dụng tính chât dấu giá trị tuyệt đối ta có 

\(\left|3-x\right|+\left|x+5\right|>=\left|3-x+x+5\right|=8\)

dấu = xảy ra <=> \(\left(3-x\right)\left(x+5\right)>=0\)

đến đây bạn tự giaỉ dấu = nhé

Cau 1: 

a: \(A=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{a-4}\)

\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}=\sqrt{a}+2\)

c: \(=\dfrac{\left|c+1\right|}{\left|c\right|-1}\)

TH1: c>0

\(C=\dfrac{c+1}{c-1}\)

TH2: c<0

\(C=\dfrac{\left|c+1\right|}{-\left(c+1\right)}=\pm1\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>=0\end{matrix}\right.\)

=>3<=x<=5

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

=>\(\sqrt{x-3}-1+\sqrt{5-x}-1=0\)

=>\(\dfrac{x-3-1}{\sqrt{x-3}+1}+\dfrac{5-x-1}{\sqrt{5-x}+1}=0\)

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

=>x-4=0

=>x=4

a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)

b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)

c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)

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