`a)`\(\left(5-3\sqrt{x}\right)\left(\sqrt{x}+9\right)=0\)
\(ĐK:x\ge0\)
\(\Leftrightarrow5-3\sqrt{x}=0\) ( vì \(\sqrt{x}+9\ge9>0\) )
\(\Leftrightarrow3\sqrt{x}=5\)
\(\Leftrightarrow x=\dfrac{25}{9}\)
Vậy \(S=\left\{\dfrac{25}{9}\right\}\)
`b)`\(3\sqrt{x}-7\sqrt{x}-10=0\)
\(ĐK:x\ge0\)
\(\Leftrightarrow-4\sqrt{x}-10=0\)
\(\Leftrightarrow-2\left(2\sqrt{x}+5\right)=0\) ( vô lý vì \(2\sqrt{x}+5\ge5>0\) )
Vậy pt vô nghiệm
`c)`\(x+\sqrt{x}-6< 0\)
\(ĐK:x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\) ( vì \(\sqrt{x}+3\ge3>0\) )
\(\Leftrightarrow x< 4\)
Vậy \(S=\left\{x|0< x< 4\right\}\)
`d)`\(\dfrac{\sqrt{x}-4}{\sqrt{x}+4}\ge0\)
\(ĐK:x\ge0\)
\(\Leftrightarrow\sqrt{x}-4\ge0\) ( vì \(\sqrt{x}+4\ge4>0\) )
\(\Leftrightarrow x\ge16\)
Vậy \(S=\left\{x|x\ge16\right\}\)
`a)(5-3\sqrt{x})(\sqrt{x}+9)=0` `ĐK: x >= 0`
Vì `\sqrt{x} >= 0<=>\sqrt{x}+9 >= 9 > 0`
`=>5-3\sqrt{x}=0`
`<=>3\sqrt{x}=5`
`<=>\sqrt{x}=5/3<=>x=25/9` (t/m)
Vậy `S={25/9}`
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`b)3\sqrt{x}-7\sqrt{x}-10=0` `ĐK: x >= 0`
`<=>-4\sqrt{x}=10`
`<=>\sqrt{x}=-5/2` (Vô lí)
Vậy ptr vô nghiệm
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`c)x+\sqrt{x}-6 < 0` `ĐK: x >= 0`
`<=>x+3\sqrt{x}-2\sqrt{x}-6 < 0`
`<=>(\sqrt{x}+3)(\sqrt{x}-2) < 0`
Với `x >= 0<=>\sqrt{x} >= 0<=>\sqrt{x}+3 >= 3 > 0`
`=>\sqrt{x}-2 < 0`
`<=>\sqrt{x} < 2`
`<=>x < 4` Kết hợp đk `x >= 0`
`=>0 <= x < 4`
Vậy `S={x|0 <= x < 4}`
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`d)[\sqrt{x}-4]/[\sqrt{x}+4] >= 0` `ĐK: x >= 0`
Với `x >= 0<=>\sqrt{x} >= 0<=>\sqrt{x}+4 >= 4 > 0`
`=>\sqrt{x}-4 >= 0`
`<=>\sqrt{x} >= 4`
`<=>x >= 16` (t/m)
Vậy `S={x|x >= 16}`
