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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) \(\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}\)
\(8,\dfrac{bc}{\sqrt{3a+bc}}=\dfrac{bc}{\sqrt{\left(a+b+c\right)a+bc}}=\dfrac{bc}{\sqrt{a^2+ab+ac+bc}}\)
\(=\dfrac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{b}{a+b}+\dfrac{c}{a+c}}{2}\)
Tương tự cho các số còn lại rồi cộng vào sẽ được
\(S\le\dfrac{3}{2}\)
Dấu "=" khi a=b=c=1
Vậy
\(7,\sqrt{\dfrac{xy}{xy+z}}=\sqrt{\dfrac{xy}{xy+z\left(x+y+z\right)}}=\sqrt{\dfrac{xy}{xy+xz+yz+z^2}}\)
\(=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{\dfrac{x}{x+z}+\dfrac{y}{y+z}}{2}\)
Cmtt rồi cộng vào ta đc đpcm
Dấu "=" khi x = y = z = 1/3
\(P=\dfrac{a^{\dfrac{1}{3}}\cdot\sqrt{b}+b^{\dfrac{1}{3}}\cdot\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}-\sqrt[3]{ab}\)
\(=\dfrac{a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{2}}+b^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{3}{6}}+a^{\dfrac{3}{6}}\cdot b^{\dfrac{2}{6}}}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=\dfrac{a^{\dfrac{2}{6}}\cdot b^{\dfrac{2}{6}}\left(a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}\right)}{a^{\dfrac{1}{6}}+b^{\dfrac{1}{6}}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
\(=a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}-a^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{3}}\)
=0
những câu tích phân như này giải tay ko hề dễ, nên mình dùng table mò ra a=13,b=18,c=78 => a+b+c=109 :v
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 \(3< 6^2\)
=> \(3^{\dfrac{1}{6}}< \left(6^2\right)^{\dfrac{1}{6}}\)
=> \(\sqrt[6]{3}< 6^{\dfrac{1}{3}}\)
=> \(\sqrt[6]{3}< \sqrt[3]{6}\)
=> \(7^{\sqrt[6]{3}}< 7^{\sqrt[3]{6}}\)
Cách làm đơn giản nhất:
Do \(\int f\left(x\right)dx=F\left(x\right)\Rightarrow F'\left(x\right)=f\left(x\right)\)
Ta có: \(F\left(x\right)=A\sqrt{1-x^3}+\dfrac{B}{1+\sqrt{x}}\)
\(\Rightarrow F'\left(x\right)=\dfrac{A\left(-3x^2\right)}{2\sqrt{1-x^3}}+B.\left(-\dfrac{\dfrac{1}{2\sqrt{x}}}{\left(1+\sqrt{x}\right)^2}\right)\)
\(\Rightarrow F'\left(x\right)=\dfrac{-3A}{2}.\dfrac{x^2}{\sqrt{1-x^3}}-\dfrac{B}{2}.\dfrac{1}{\sqrt{x}\left(1+\sqrt{x}\right)^2}=f\left(x\right)\)
Đồng nhất hệ số ta được:
\(\left\{{}\begin{matrix}\dfrac{-3A}{2}=1\\\dfrac{-B}{2}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}A=\dfrac{-2}{3}\\B=-2\end{matrix}\right.\) \(\Rightarrow A+B=-\dfrac{8}{3}\)
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) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) =
= 
= 
= 