Toán Lớp 9: 1/ Tính.
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√53 ²-28 ² ; 2/ √11 -3 – √11
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√4-2 √3 – √ 4-2 √3
2/ Giải pt.
a/ x-5 √x + x =0
b/ ———- ———
√ 2x-1 = √ 3x-5
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1)\sqrt {{{53}^2} – {{28}^2}} \\
= \sqrt {\left( {53 – 28} \right).\left( {53 + 28} \right)} \\
= \sqrt {25.81} \\
= 5.9\\
= 45\\
2)\sqrt {11} – 3 – \sqrt {11} = – 3\\
3)\sqrt {4 – 2\sqrt 3 } = \sqrt {3 – 2.\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} \\
= \sqrt 3 – 1\\
B2)\\
a)Dkxd:x \ge 0\\
x – 5\sqrt x + x = 0\\
\Leftrightarrow 2x – 5\sqrt x = 0\\
\Leftrightarrow \sqrt x \left( {2\sqrt x – 5} \right) = 0\\
\Leftrightarrow x = 0;x = \dfrac{{25}}{4}\\
Vậy\,x = 0;x = \dfrac{{25}}{4}\\
b)Dkxd:\left\{ \begin{array}{l}
2x – 1 \ge 0\\
3x – 5 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \ge \dfrac{5}{2}
\end{array} \right. \Leftrightarrow x \ge \dfrac{5}{2}\\
\sqrt {2x – 1} = \sqrt {3x – 5} \\
\Leftrightarrow 2x – 1 = 3x – 5\\
\Leftrightarrow x = 4\left( {tm} \right)\\
Vậy\,x = 4
\end{array}$