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![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\dfrac{a^{\dfrac{1}{3}}\cdot\sqrt{b}+b^{\dfrac{1}{3}}\cdot\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}-\sqrt[3]{ab}\)
\(=\dfrac{a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{2}}+b^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{3}{6}}+a^{\dfrac{3}{6}}\cdot b^{\dfrac{2}{6}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{2}{6}}\left(a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}\right)}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(α) chứa AB và song song với CD
⇒ (α) nhận (1; 0; -1) là 1 vtpt
(α) đi qua A(-2; 6; 3)
⇒ (α): x – z + 5 = 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\left(2+\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\right)=\frac{\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)^2}{\sqrt[3]{ab}}:\frac{2\sqrt[3]{ab}+\left(\sqrt[3]{a}\right)^2+\left(\sqrt[3]{a}\right)^2}{\sqrt[3]{ab}}\)
\(=\frac{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}{\sqrt[3]{ab}}-\frac{\sqrt[3]{ab}}{\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^2}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left[\frac{\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\left(a+a^{\frac{1}{2}}b^{\frac{1}{2}}+b\right)}{a^{\frac{1}{2}}-b^{\frac{1}{2}}}+a^{\frac{1}{2}}b^{\frac{1}{2}}\right]\left[\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}\right]^2\)
\(=\frac{a+2a^{\frac{1}{2}}b^{\frac{1}{2}}+b}{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^2}=\frac{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^2}{\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)^2}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
Chọn A.
Ta có:![](http://cdn.hoc24.vn/bk/Sv0y9mSZZZq7.png)