Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
\(I=\frac{a^{\frac{4}{3}}-8a^{\frac{2}{3}}b}{a^{\frac{2}{3}}+2\sqrt[3]{ab}+4b^{\frac{2}{3}}}\left(1-2\sqrt[3]{\frac{b}{a}}\right)^{-1}-a^{\frac{2}{3}}=\frac{a^{\frac{1}{3}}\left(a-8b\right)}{a^{\frac{2}{3}}+2a^{\frac{1}{3}}.b^{\frac{1}{3}}+4b^{\frac{2}{3}}}\left(\frac{\sqrt[3]{a}-2\sqrt[3]{b}}{\sqrt[3]{a}}\right)^{-1}-a^{\frac{2}{3}}\)
\(=\frac{\sqrt[3]{a}\left[\left(\sqrt[3]{a}\right)^3-\left(2\sqrt[3]{b}\right)^3\right]}{a^{\frac{2}{3}}+2\sqrt[3]{ab}+4b^{\frac{2}{3}}}.\frac{\sqrt[3]{a}}{\sqrt[3]{a}-2\sqrt[3]{b}}-a^{\frac{2}{3}}\)
\(=\frac{\left(\sqrt[3]{a}\right)^2\left(\sqrt[3]{a}-2\sqrt[3]{b}\right)\left[\left(\sqrt[3]{a}\right)^2+2\sqrt[3]{ab}+\left(2\sqrt[3]{b}\right)^2\right]}{\left(\sqrt[3]{a}-a\sqrt[3]{b}\right)\left[\left(\sqrt[3]{a}\right)^2+2\sqrt[3]{ab}+\left(2\sqrt[3]{b}\right)^2\right]}-a^{\frac{2}{3}}=a^{\frac{2}{3}}-a^{\frac{2}{3}}=0\)
Câu a, b thì Nguyễn Quang Duy làm đúng rồi.
c) \(a^{\dfrac{4}{3}}:\sqrt[3]{a}=a^{\dfrac{4}{3}}:a^{\dfrac{1}{3}}=a^{\dfrac{4}{3}-\dfrac{1}{3}}=a\)
d) \(\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}\)
Theo công thức biến đổi có số ta có : \(\log_{a^n}x=\frac{\log_ax}{\log_aa^n}=\frac{1}{n}\log_ax\)
Từ đó ta có :
\(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}\)
\(=\frac{1}{\log_ax}+\frac{2}{\log_ax}+\frac{4}{\log_ax}+...+\frac{n}{\log_ax}\)
\(=\frac{1+2+3+...+n}{\log_ax}=\frac{n\left(n+1\right)}{\log_ax}\)
Vậy \(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}=\frac{n\left(n+1\right)}{\log_ax}\)
Lời giải:
Ta có \(A=\frac{a^{\frac{1}{3}}-a^{\frac{7}{3}}}{a^{\frac{1}{3}}-a^{\frac{4}{3}}}-\frac{a^{\frac{1}{3}}-a^{\frac{5}{3}}}{a^{\frac{2}{3}}+a^{\frac{1}{3}}}\)
\(=\frac{\sqrt[3]{a}-\sqrt[3]{a^7}}{\sqrt[3]{a}-\sqrt[3]{a^4}}-\frac{\sqrt[3]{a}-\sqrt[3]{a^5}}{\sqrt[3]{a^2}+\sqrt[3]{a}}\)
\(=\frac{\sqrt[3]{a}(1-a^2)}{\sqrt[3]{a}(1-a)}-\frac{\sqrt[3]{a}(1-\sqrt[3]{a^4})}{\sqrt[3]{a}(1+\sqrt[3]{a})}=\frac{1-a^2}{1-a}-\frac{1-\sqrt[3]{a^4}}{1+\sqrt[3]{a}}\)
\(=1+a-\frac{1-\sqrt[3]{a^4}}{1+\sqrt[3]{a}}\)
Đặt \(\sqrt[3]{a}=t\Rightarrow A=1+t^3-\frac{1-t^4}{1+t}=1+t^3-\frac{(1-t^2)(1+t^2)}{1+t}\)
\(=1+t^3-\frac{(1-t)(1+t)(1+t^2)}{1+t}=1+t^3-(1-t)(1+t^2)\)
\(=2t^3-t^2+t\)
4 a - 9 a - 1 2 a 1 2 - 3 a - 1 2 + a - 4 + 3 a - 1 a 1 2 - a - 1 2
Chọn B