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\(f'\left(x\right)=1-\dfrac{2x}{\sqrt{x^2+12}}\le0\\ \Leftrightarrow\sqrt{x^2+12}\le2x\\ \Leftrightarrow\left\{{}\begin{matrix}x^2+12\le4x^2\\x\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x^2\ge12\\x\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2\ge4\\x\ge0\end{matrix}\right.\Leftrightarrow x\ge2\)
Đáp số : \(\left[2,+\infty\right]\)
Lời giải:
a) Ta có f'(x) = 3x2 + 1, g(x) = 6x + 1. Do đó
f'(x) > g'(x) <=> 3x2 + 1 > 6x + 1 <=> 3x2 - 6x >0
<=> 3x(x - 2) > 0 <=> x > 2 hoặc x > 0 <=> x ∈ (-∞;0) ∪ (2;+∞).
b) Ta có f'(x) = 6x2 - 2x, g'(x) = 3x2 + x. Do đó
f'(x) > g'(x) <=> 6x2 - 2x > 3x2 + x <=> 3x2 - 3x > 0
<=> 3x(x - 1) > 0 <=> x > 1 hoặc x < 0 <=> x ∈ (-∞;0) ∪ (1;+∞).
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
