Toán Lớp 9: A=(1/√x – 1/√x-1 ) ÷ ( √x + 2 / √x-1 – √x+1/√x-2)
a, Rút gọn A
b, Tìm x để A = 0
Gạc xược là phần nha . Kiểu ngăn giữa tử vs mẫu í . q1
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Toán Lớp 9: A=(1/√x – 1/√x-1 ) ÷ ( √x + 2 / √x-1 – √x+1/√x-2)
a, Rút gọn A
b, Tìm x để A = 0
Gạc xược là phần nha . Kiểu ngăn giữa tử vs mẫu í . q1
Comments ( 1 )
a)Dkxd:x > 0;x\# 1;x\# 4\\
A = \left( {\dfrac{1}{{\sqrt x }} – \dfrac{1}{{\sqrt x – 1}}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x – 1}} – \dfrac{{\sqrt x + 1}}{{\sqrt x – 2}}} \right)\\
= \dfrac{{\sqrt x – 1 – \sqrt x }}{{\sqrt x \left( {\sqrt x – 1} \right)}}:\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) – \left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{ – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)}}{{x – 4 – x + 1}}\\
= \dfrac{{ – 1}}{{\sqrt x }}.\dfrac{{\sqrt x – 2}}{{ – 3}}\\
= \dfrac{{\sqrt x – 2}}{{3\sqrt x }}\\
b)A = 0\\
\Leftrightarrow \dfrac{{\sqrt x – 2}}{{3\sqrt x }} = 0\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {ktm} \right)
\end{array}$