Bài 1: Căn bậc hai

LT

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A=\(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)

TK
1 tháng 9 2017 lúc 6:18

\(A=\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}=\dfrac{2a^2+4}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}=\dfrac{2a^2+4}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)

\(=\dfrac{2a^2+4}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}=\dfrac{2a^2+4-\left(1+a+a^2\right)\left[\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)\right]}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}=\dfrac{2a^2+4-2\left(1+a+a^2\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}\)

\(=\dfrac{2a^2+4-2-2a-2a^2}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}=\dfrac{2-2a}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{2}{1+a+a^2}\)

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LT
1 tháng 9 2017 lúc 9:13

A=\(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)

=\(\dfrac{2a^2+4}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)=

\(\dfrac{2a^2+4}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}\)

= \(\dfrac{2a^2+4-\left(1-\sqrt{a}+a-a\sqrt{a}+a^2-a^2\sqrt{a}\right)-\left(1+\sqrt{a}+a+a\sqrt{a}+a^2+a^2\sqrt{2}\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}\)

=\(\dfrac{-2a-2}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{-2}{1+a+a^2}\)

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