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bài 2: ta có : \(Q=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-\left(1-a\right)}\right)\left(\sqrt{\dfrac{1}{a^2}-1}-\dfrac{1}{a}\right).\sqrt{a^2-2a+1}\)
\(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}\sqrt{1-a}+1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\dfrac{\sqrt{1-a^2}}{a}-\dfrac{1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1-a^2}+1}{a}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{1-a^2-1}{a^2}\right)\left(1-a\right)=a-1\)b) ta có : \(Q^3-Q=\left(a-1\right)\left(\left(a-1\right)^2-1\right)=a\left(a-1\right)\left(a-2\right)\)
mà ta có : \(\left\{{}\begin{matrix}a>0\\a-1< 0\\a-2< 0\end{matrix}\right.\Rightarrow a\left(a-1\right)\left(a-2\right)>0\) \(\Rightarrow Q^3-Q>0\Leftrightarrow Q^3>Q\)
vậy \(Q^3>Q\)
Nguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh Hằngsoyeon_Tiểubàng giảiMashiro ShiinaVõ Đông Anh Tuấn
Hoàng Lê Bảo NgọcTrần Việt Linh
cứu tôi với
1)
DKCĐ: a>0,\(a\ne1\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)
Câu b nhé:
Ta có:
\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)
Trước hết ta có:
\(23^{2018}+23^{2020}>2\sqrt{23^{2018}.23^{2020}}=2\sqrt{23^{4038}}=2.23^{2019}\)
Dễ dàng nhận ra \(A>0\) và \(B>0;\) xét thương:
\(\dfrac{A}{B}=\dfrac{23^{2018}+1}{23^{2019}+1}\div\dfrac{23^{2019}+1}{23^{2020}+1}=\dfrac{\left(23^{2018}+1\right)\left(23^{2020}+1\right)}{\left(23^{2019}+1\right)^2}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{23^{4038}+23^{2018}+23^{2020}+1}{\left(23^{2019}+1\right)^2}=\dfrac{\left(23^{2019}\right)^2+23^{2018}+23^{2020}+1}{\left(23^{2019}+1\right)^2}\)
\(\Rightarrow\dfrac{A}{B}>\dfrac{\left(23^{2019}\right)^2+2.23^{2019}+1}{\left(23^{2019}+1\right)^2}=\dfrac{\left(23^{2019}+1\right)^2}{\left(23^{2019}+1\right)^2}=1\)
\(\Rightarrow\dfrac{A}{B}>1\Rightarrow A>B\)