\(\sqrt{5x}-2\le4\Rightarrow\sqrt{5x}\le6.\) 

I5xI<=36

\(\orbr{\begin{cases}x< =\frac{36}{5}\approx7^+\\x>=\frac{-36}{5}\approx7^-\end{cases}}\),

S={-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7)

\(\sqrt{5x}-2< =4\)                                       ĐK:\(\sqrt{5x}>0\)<=> 5x > 0 <=> x>0

<=>\(\sqrt{5x}< =4+2\)

<=>\(\sqrt{5x}\)<= 6

<=> 5x <= \(6^2\)

<=>5x <= 36

<=> x <= \(\frac{36}{5}\)

<=> x <= 7,2

\(\sqrt{5x-2}\le4\)

<=>\(\begin{cases}5x-2\ge0\\5x-2\le16\end{cases}\)<=> \(\begin{cases}x\ge\frac{2}{5}\\x\le\frac{18}{5}\end{cases}\)

<=>x=1,2,3

\(\sqrt{\frac{3x-1}{x+2}}=\sqrt{5}\)

<=> \(\begin{cases}\frac{3x-1}{x+2}\ge0\\3x-1=5x+10\end{cases}\)

=> x=-11/2

thay x=-11/2 vào \(\frac{3x-1}{x+2}\)>=0 thỏa 

=> nghiệm bpt là x=-11/2

\(pt\Leftrightarrow\sqrt{x^2+x-6}=\sqrt{x^2+2}\)

Ta thấy 2 vế luôn dương bình phương lên ta có:

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

\(\Rightarrow x^2+x-6=x^2+2\)

\(\Rightarrow x^2-x^2+x=6+2\)

\(\Rightarrow x=8\)

\(\sqrt{x-1}\cdot\sqrt{\frac{x+1}{x-1}}=6-\sqrt{x+1}\left(ĐK:x\ge1\right)\)

\(\Leftrightarrow\sqrt{x+1}=6-\sqrt{x+1}\)

\(\Leftrightarrow2\sqrt{x+1}=6\)

\(\Leftrightarrow\sqrt{x+1}=3\)

\(\Leftrightarrow x+1=9\)

\(\Leftrightarrow x=8\)

ƯC(8;12) = { -4; - 2; -1 ; 1 ;  2; 4 }

ƯC(8,12)={-4;-2;-1;1;2;4}

{ 41;82 }

tick mình nha!!