Toán Lớp 9: Giúp mik vs ạ
Please????????
1) tìm x : 2×2√x-x=3
2) a) (1/√3 -√2 + √3 -√6/1-√2 -√27)×√5+2√6
b) (2+x+√x/1+√x)×(2- x-√x/1-√x)
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Toán Lớp 9: Giúp mik vs ạ
Please????????
1) tìm x : 2×2√x-x=3
2) a) (1/√3 -√2 + √3 -√6/1-√2 -√27)×√5+2√6
b) (2+x+√x/1+√x)×(2- x-√x/1-√x)
Comments ( 1 )
1)Dkxd:x \ge 0\\
2.2\sqrt x – x = 3\\
\Leftrightarrow x – 4\sqrt x + 3 = 0\\
\Leftrightarrow x – \sqrt x – 3\sqrt x + 3 = 0\\
\Leftrightarrow \left( {\sqrt x – 1} \right)\left( {\sqrt x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 9
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,x = 1;x = 9\\
2)a)\left( {\dfrac{1}{{\sqrt 3 – \sqrt 2 }} + \dfrac{{\sqrt 3 – \sqrt 6 }}{{1 – \sqrt 2 }} – \sqrt {27} } \right).\sqrt {5 + 2\sqrt 6 } \\
= \left( {\dfrac{{\sqrt 3 + \sqrt 2 }}{{3 – 2}} + \dfrac{{\sqrt 3 \left( {1 – \sqrt 2 } \right)}}{{1 – \sqrt 2 }} – 3\sqrt 3 } \right).\sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} \\
= \left( {\sqrt 3 + \sqrt 2 + \sqrt 3 – 3\sqrt 3 } \right).\left( {\sqrt 3 + \sqrt 2 } \right)\\
= \left( {\sqrt 2 – \sqrt 3 } \right).\left( {\sqrt 3 + \sqrt 2 } \right)\\
= 2 – 3\\
= – 1\\
b)\left( {2 + \dfrac{{x + \sqrt x }}{{1 + \sqrt x }}} \right).\left( {2 – \dfrac{{x – \sqrt x }}{{1 – \sqrt x }}} \right)\\
= \dfrac{{2 + 2\sqrt x + x + \sqrt x }}{{1 + \sqrt x }}.\dfrac{{2 – 2\sqrt x – x + \sqrt x }}{{1 – \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{1 + \sqrt x }}.\dfrac{{\left( {1 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}{{1 – \sqrt x }}\\
= \left( {\sqrt x + 2} \right)\left( {2 + \sqrt x } \right)\\
= x + 4\sqrt x + 4
\end{array}$