Toán Lớp 7: a,√x+3-2=0
b, √5x-1=2
c, √0,81[ √x+ √16/144]=9/10
Tìm x giúp mình với ha
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Toán Lớp 7: a,√x+3-2=0
b, √5x-1=2
c, √0,81[ √x+ √16/144]=9/10
Tìm x giúp mình với ha
Comments ( 1 )
a)Dkxd:x \ge – 3\\
\sqrt {x + 3} – 2 = 0\\
\Leftrightarrow \sqrt {x + 3} = 2\\
\Leftrightarrow x + 3 = 4\\
\Leftrightarrow x = 1\left( {tm} \right)\\
Vậy\,x = 1\\
b)Dkxd:x \ge \dfrac{1}{5}\\
\sqrt {5x – 1} = 2\\
\Leftrightarrow 5x – 1 = 4\\
\Leftrightarrow 5x = 5\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1\\
c)Dkxd:x \ge 0\\
\sqrt {0,81} .\left( {\sqrt x + \sqrt {\dfrac{{16}}{{144}}} } \right) = \dfrac{9}{{10}}\\
\Leftrightarrow 0,9.\left( {\sqrt x + \dfrac{4}{{12}}} \right) = 0,9\\
\Leftrightarrow \sqrt x + \dfrac{1}{3} = 1\\
\Leftrightarrow \sqrt x = \dfrac{2}{3}\\
\Leftrightarrow x = \dfrac{4}{9}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{4}{9}\\
d)Dkxd:x \ge – 1\\
\left| {\dfrac{1}{3}\sqrt {x + 1} – \dfrac{2}{9}} \right| – \dfrac{1}{6} = \dfrac{1}{9}\\
\Leftrightarrow \left| {\dfrac{1}{3}\sqrt {x + 1} – \dfrac{2}{9}} \right| = \dfrac{1}{9} + \dfrac{1}{6} = \dfrac{5}{{18}}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{3}\sqrt {x + 1} – \dfrac{2}{9} = \dfrac{5}{{18}}\\
\dfrac{1}{3}\sqrt {x + 1} – \dfrac{2}{9} = – \dfrac{5}{{18}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{3}\sqrt {x + 1} = \dfrac{5}{{18}} + \dfrac{2}{9} = \dfrac{1}{2}\\
\dfrac{1}{3}\sqrt {x + 1} = – \dfrac{5}{{18}} + \dfrac{2}{9} = \dfrac{{ – 1}}{{18}}\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow \sqrt {x + 1} = \dfrac{3}{2}\\
\Leftrightarrow x + 1 = \dfrac{9}{4}\\
\Leftrightarrow x = \dfrac{5}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{5}{4}
\end{array}$