a/\(\sqrt{x}=7\)
\(\Leftrightarrow x=49\)
b/\(\Leftrightarrow x< 4\)(do x>0)
\(\Rightarrow x\varepsilon\left\{0;1;2;3\right\}\)
c/\(2x< 16\)
\(\Leftrightarrow x< 8\)
\(\Leftrightarrow x\varepsilon\left\{1;2;3;4;5;6;7\right\}\)
a) \(2\sqrt{x}=14\Leftrightarrow\sqrt{x}=7\)
\(\Leftrightarrow x=7^2\Leftrightarrow x=49\)
b) \(\sqrt{x}< \sqrt{2}\Leftrightarrow x< 2\)
c) \(\sqrt{2x}< 4\)
Vì \(4=\sqrt{16}\text{ nên }\sqrt{2x}< 4\text{ có nghĩa là }\sqrt{2x}< 16\)
\(\Leftrightarrow2x< 16\)
\(\Leftrightarrow x< 8\left(x\ge0\right)\)
a) \(2\sqrt{x}=14\)
\(\Rightarrow\sqrt{x}=\frac{14}{2}=7\)
\(\Rightarrow x=49\)
b) ...
a) \(2\sqrt{x}=14\)
\(\sqrt{x}=14:2=7\)
=> x = 7^2 = 49
b ) \(\sqrt{x}< \sqrt{2}\)
=> \(0\le x< 2\)( do x ko am )
=> x \(\in\left\{0;1\right\}\)
c )\(\sqrt{2x}< 4\)
=> 2x < 4^2
=> 2x < 16
=> x < 8
Vi x ko am ( hay x la so nguyen duong )
=> \(x\in\left\{0;1;2;3;4;5;6;7\right\}\)
a) \(2\sqrt{x}=14\)( ĐKXĐ : \(x\ge0\))
<=> \(\sqrt{x}=7\)
Bình phương hai vế
<=> \(\left(\sqrt{x}\right)^2=7^2\)
<=> \(x=49\)( tmđk )
b) \(\sqrt{x}< \sqrt{2}\) ( ĐKXĐ : \(x\ge0\))
Bình phương hai vế
<=> \(\left(\sqrt{x}\right)^2< \left(\sqrt{2}\right)^2\)
<=> \(x< 2\)
Kết hợp ĐKXĐ => \(0\le x< 2\)
c) \(\sqrt{2x}< 4\)( ĐKXĐ : \(x\ge0\))
Bình phương hai vế
<=> \(\left(\sqrt{2x}\right)^2< 4^2\)
<=> \(2x< 16\)
<=> \(x< 8\)
Kết hợp ĐKXĐ => \(0\le x< 8\)
a)
\(2\sqrt{x}=14\)
\(\sqrt{x}=7\)
\(\hept{\begin{cases}7\ge0\left(llđ\right)\\x=7^2\end{cases}}\)
\(x=49\)
b)
\(\sqrt{x}< \sqrt{2}\)
\(\hept{\begin{cases}x\ge0\\x< 2\end{cases}}\)
\(\Rightarrow0\le x< 2\)
c)
\(\sqrt{2x}< 4\)
\(\hept{\begin{cases}2x\ge0\\4>0\left(llđ\right)\\2x< 4^2\end{cases}}\)
\(\hept{\begin{cases}x\ge0\\x< 8\end{cases}}\)
\(\Rightarrow0\le x< 8\)
