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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) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
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) \(\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}\)
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) (ĐK: \(x\ge0;x\ne\dfrac{1}{9}\))
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}\right)^2-1^2}\right]:\left[\dfrac{\left(3\sqrt{x}+1\right)\cdot1}{3\sqrt{x}+1}-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right]\)
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}\)
\(A=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3x+3\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)
\(A=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
Lời giải:
a)
\(\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
b)
\(\left(3-\frac{\sqrt{2}+2}{\sqrt{2}+1}\right)\left(3+\frac{2-\sqrt{2}}{\sqrt{2}-1}\right)=\left(3-\frac{\sqrt{2}(1+\sqrt{2})}{\sqrt{2}+1}\right)\left(3+\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}\right)\)
\(=(3-\sqrt{2})(3+\sqrt{2})=3^2-(\sqrt{2})^2=9-2=7\)
a) ta có 2√5= = √20 ; 3√2 =
= √ 18 => 2√5 > 3√2
=> <
b) 6√3 = = √108 ; 3√6 =
= √54 => 6√3 > 3√6 =>
>
a) \(2\sqrt{5}=\sqrt{2^2.5}=\sqrt{20}\)
\(3\sqrt{2}=\sqrt{3^2.2}=\sqrt{18}\)
=> \(2\sqrt{5}>3\sqrt{2}\)
=> \(\left(\dfrac{1}{3}\right)^{2\sqrt{5}}< \left(\dfrac{1}{3}\right)^{3\sqrt{2}}\)
(vì cơ số \(\dfrac{1}{3}< 1\))
b) Vì \(3< 6^2\)
=> \(3^{\dfrac{1}{6}}< \left(6^2\right)^{\dfrac{1}{6}}\)
=> \(\sqrt[6]{3}< 6^{\dfrac{1}{3}}\)
=> \(\sqrt[6]{3}< \sqrt[3]{6}\)
=> \(7^{\sqrt[6]{3}}< 7^{\sqrt[3]{6}}\)
\(P=\dfrac{a^{\dfrac{1}{3}}\cdot\sqrt{b}+b^{\dfrac{1}{3}}\cdot\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}-\sqrt[3]{ab}\)
\(=\dfrac{a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{2}}+b^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{3}{6}}+a^{\dfrac{3}{6}}\cdot b^{\dfrac{2}{6}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{2}{6}}\left(a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}\right)}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
=0