Rút gọn:α,√15-√5/1-√5 b, x^2-3/x+√3 c, √x +2/x-4 d,x+√5/x^2+2√5x+5 16/07/2021 Bởi Everleigh Rút gọn:α,√15-√5/1-√5 b, x^2-3/x+√3 c, √x +2/x-4 d,x+√5/x^2+2√5x+5
Đáp án: b. \(x – \sqrt 3 \) Giải thích các bước giải: \(\begin{array}{l}a.\dfrac{{\sqrt {15} – \sqrt 5 }}{{1 – \sqrt 5 }} = \dfrac{{\sqrt 5 \left( {\sqrt 3 – 1} \right)}}{{1 – \sqrt 5 }}\\b.\dfrac{{{x^2} – 3}}{{x + \sqrt 3 }} = \dfrac{{\left( {x + \sqrt 3 } \right)\left( {x – \sqrt 3 } \right)}}{{x + \sqrt 3 }}\\ = x – \sqrt 3 \\c.\dfrac{{\sqrt x + 2}}{{x – 4}} = \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{1}{{\sqrt x – 2}}\\d.\dfrac{{x + \sqrt 5 }}{{{x^2} + 2\sqrt 5 .x + 5}} = \dfrac{{x + \sqrt 5 }}{{{{\left( {x + \sqrt 5 } \right)}^2}}}\\ = \dfrac{1}{{x + \sqrt 5 }}\end{array}\) Bình luận
Đáp án:
b. \(x – \sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{\sqrt {15} – \sqrt 5 }}{{1 – \sqrt 5 }} = \dfrac{{\sqrt 5 \left( {\sqrt 3 – 1} \right)}}{{1 – \sqrt 5 }}\\
b.\dfrac{{{x^2} – 3}}{{x + \sqrt 3 }} = \dfrac{{\left( {x + \sqrt 3 } \right)\left( {x – \sqrt 3 } \right)}}{{x + \sqrt 3 }}\\
= x – \sqrt 3 \\
c.\dfrac{{\sqrt x + 2}}{{x – 4}} = \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{1}{{\sqrt x – 2}}\\
d.\dfrac{{x + \sqrt 5 }}{{{x^2} + 2\sqrt 5 .x + 5}} = \dfrac{{x + \sqrt 5 }}{{{{\left( {x + \sqrt 5 } \right)}^2}}}\\
= \dfrac{1}{{x + \sqrt 5 }}
\end{array}\)