So sánh \(\frac{46-3\sqrt{39}}{4}\) và \(\sqrt{50}\)
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a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)
\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)
Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)
b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)
\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)
Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)
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B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)
\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)
\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)
\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))
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Xét phân thức phụ sau, với n nguyên dương lớn hơn 1 ta có:
Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{2\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}\right)^2\sqrt{n}}=2\left(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}\right)\sqrt{n}}\right)\)
\(=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán ta được:
\(A=2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\right)\)
\(A=2-\frac{2}{\sqrt{2020}}< 2=B\)
Vậy A < B
Ta có:
\(\frac{46-3\sqrt{39}}{4}< \frac{46-3\sqrt{36}}{4}=\frac{46-18}{4}=7=\sqrt{49}< \sqrt{50}.\)