Toán Lớp 9: Giúp em vs ạ em cần nộp gấp ai đầu em vote 5 sao ạ
C = √x + 2/√x + 3 + √x/ 2 – √x + x + 2√x – 8/(√x + 3)(√x -2)
a,tìm điều kiện xác định
b,rút gọn C
c, tìm x để c<1
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: Giúp em vs ạ em cần nộp gấp ai đầu em vote 5 sao ạ
C = √x + 2/√x + 3 + √x/ 2 – √x + x + 2√x – 8/(√x + 3)(√x -2)
a,tìm điều kiện xác định
b,rút gọn C
c, tìm x để c<1
Comments ( 1 )
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x – 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
b)\\
C = \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{2 – \sqrt x }} + \dfrac{{x + 2\sqrt x – 8}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) – \sqrt x \left( {\sqrt x + 3} \right) + x + 2\sqrt x – 8}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x – 4 – x – 3\sqrt x + x + 2\sqrt x – 8}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x – \sqrt x – 12}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x – 4} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x – 4}}{{\sqrt x – 2}}\\
c)C < 1\\
\Leftrightarrow \dfrac{{\sqrt x – 4}}{{\sqrt x – 2}} – 1 < 0\\
\Leftrightarrow \dfrac{{\sqrt x – 4 – \left( {\sqrt x – 2} \right)}}{{\sqrt x – 2}} < 0\\
\Leftrightarrow \dfrac{{\sqrt x – 4 – \sqrt x + 2}}{{\sqrt x – 2}} < 0\\
\Leftrightarrow \dfrac{{ – 2}}{{\sqrt x – 2}} < 0\\
\Leftrightarrow \sqrt x – 2 > 0\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > 4\\
Vậy\,x > 4
\end{array}$