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1/
Ta có: \(\left(1+\sqrt{15}\right)^2\)= 1 + 15 + \(2\sqrt{15}\)= 16 + \(2\sqrt{15}\)
\(\sqrt{24}^2\)= 24 = 16 + 8
Vì: \(\sqrt{15}^2\)= 15 < 16 =\(4^2\)
Nên: \(\sqrt{15}< 4\)
=> \(2\sqrt{15}< 8\)
=> \(16+2\sqrt{15}< 24\)
=> \(\left(1+\sqrt{15}\right)^2< \sqrt{24}^2\)
Vậy \(1+\sqrt{15}< \sqrt{24}\)
2/
b/ \(3x-7\sqrt{x}=20\)\(\left(x\ge0\right)\)
<=> \(3x-7\sqrt{x}-20=0\)
<=> \(3x-12\sqrt{x}+5\sqrt{x}-20=0\)
<=> \(3\sqrt{x}\left(\sqrt{x}-4\right)+5\left(\sqrt{x}-4\right)=0\)
<=> \(\left(\sqrt{x}-4\right)\left(3\sqrt{x}+5\right)=0\)
<=> \(\sqrt{x}-4=0\)hoặc \(3\sqrt{x}+5=0\)
<=> \(\sqrt{x}=4\)hoặc \(3\sqrt{x}=-5\)(vô nghiệm)
<=> \(x=16\)
Vậy S=\(\left\{16\right\}\)
c/ \(1+\sqrt{3x}>3\)
<=> \(\sqrt{3x}>2\)
<=> \(3x>4\)
<=> \(x>\frac{4}{3}\)
d/ \(x^2-x\sqrt{x}-5x-\sqrt{x}-6=0\)(\(x\ge0\))
<=> \(\left(x^2-5x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)
<=> \(\left(x^2-6x+x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)
<=> \([x\left(x-6\right)+\left(x-6\right)]-\sqrt{x}\left(x+1\right)=0\)
<=> \(\left(x-6\right)\left(x+1\right)-\sqrt{x}\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(x-6-\sqrt{x}\right)=0\)
<=> \(\left(x+1\right)\left(x-3\sqrt{x}+2\sqrt{x}-6\right)=0\)
<=> \(\left(x+1\right)[\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)]=0\)
<=> \(\left(x+1\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)
<=> \(x+1=0\) hoặc \(\sqrt{x}-3=0\)hoặc \(\sqrt{x}+2=0\)
<=> \(x=-1\)(loại) hoặc \(x=9\)hoặc \(\sqrt{x}=-2\)(vô nghiệm)
Vậy S={ 9 }
1) \(\sqrt{2019}+1>\sqrt{1936}+1=44+1=45>44\)
2) \(\sqrt{\left(\sqrt{2}-1\right)^2}+3\sqrt{2}=\sqrt{2}-1+3\sqrt{2}=4\sqrt{2}-1\)
3) \(\sqrt{a^2b^6}=\sqrt{a^2\cdot\left(b^3\right)^2}=a\cdot\left|b^3\right|=-a\cdot b^3\)
4) \(3\sqrt{x}\ge7\Leftrightarrow\sqrt{x}\ge\frac{7}{3}\Leftrightarrow x\ge\frac{49}{9}\)