Toán Lớp 9: cho biết A= x-√x/x-9 + 1/√x+3 – 1/√x-3 ( x ≥ 0, x khác 9)
a. rút gọn A
b.tính giá trị biểu thức A khi x = 16
c. tìm x , để A=3/4
d. tìm x thuộc Z để A thuộc Z
giúp mk vs mk đang cần gấp lắm mk hứa cho 5 sao nếu trả lời
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a)Dkxd:x \ge 0;x \ne 9\\
A = \dfrac{{x – \sqrt x }}{{x – 9}} + \dfrac{1}{{\sqrt x + 3}} – \dfrac{1}{{\sqrt x – 3}}\\
= \dfrac{{x – \sqrt x + \sqrt x – 3 – \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{x – 3 – \sqrt x – 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{x – \sqrt x – 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}}\\
b)x = 16\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow A = \dfrac{{4 + 2}}{{4 + 3}} = \dfrac{6}{7}\\
c)A = \dfrac{3}{4}\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} = \dfrac{3}{4}\\
\Leftrightarrow 4\left( {\sqrt x + 2} \right) = 3\left( {\sqrt x + 3} \right)\\
\Leftrightarrow 4\sqrt x + 8 = 3\sqrt x + 9\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1\\
d)A = \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 – 1}}{{\sqrt x + 3}}\\
= 1 – \dfrac{1}{{\sqrt x + 3}}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 3 = 1\\
\sqrt x + 3 = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = – 2\left( {ktm} \right)\\
\sqrt x = – 4\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x \in \emptyset
\end{array}$