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HT
20 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}x^2\left(1+\dfrac{x}{x^2}-\dfrac{1}{x^2}\right)=+\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}+\lim\limits_{x\rightarrow+\infty}2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}+2\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}}+\dfrac{x}{x}}=\dfrac{1}{2}+\dfrac{2}{2}=\dfrac{3}{2}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2-x}{\sqrt{x^2+x}+x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x^2}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{2x}{x^2}+\dfrac{x}{x^2}}}+2\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{x^2}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}=0\)
13 tháng 3 2021
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt{8x^2+5x+2}}=\dfrac{1+\sqrt{1+\dfrac{2}{x^2}}}{\sqrt{8+\dfrac{5}{x}+\dfrac{2}{x^2}}}=\dfrac{1+\sqrt{1}}{\sqrt{8}}=\dfrac{\sqrt{2}}{2}\).
13 tháng 3 2021
Thiếu \(\lim\limits_{x\rightarrow-\infty}\) ở sau dấu bằng thứ nhất nha
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 4 2022
\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)
\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
HT
8 tháng 4 2021
1/ \(y'=\dfrac{\left(\sqrt{x+1}\right)'x-x'\sqrt{x+1}}{x^2}=\dfrac{\dfrac{x}{2\sqrt{x+1}}-\sqrt{x+1}}{x^2}=\dfrac{-x-2}{2x^2\sqrt{x+1}}\)
2/ \(y'=\dfrac{1-x^2-\left(1-x^2\right)'x}{\left(1-x^2\right)^2}=\dfrac{1+x^2}{\left(1-x^2\right)^2}\)
3/ \(y'=\dfrac{-\left(x-\sqrt{x+1}\right)'}{\left(x-\sqrt{x+1}\right)^2}=\dfrac{-1+\dfrac{1}{2\sqrt{x+1}}}{\left(x-\sqrt{x+1}\right)^2}\)
4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)
\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)
5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học
AM
9 tháng 2 2022
a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)
b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)
c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)
d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe
HT
20 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2-x-1}{\sqrt{x^2-x+1}+\sqrt{x^2+x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2x}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}=-\dfrac{2}{1+1}=-1\)
b/ \(=\lim\limits_{x\rightarrow2}\dfrac{4x+1-9}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\dfrac{4}{\left(2+2\right)\left(\sqrt{4.2+1}+3\right)}=\dfrac{1}{6}\)
c/ \(=\lim\limits_{x\rightarrow-2}\dfrac{2x+5-1}{\left(x-2\right)\left(x+2\right)\left(\sqrt{2x+5}+1\right)}=\lim\limits_{x\rightarrow-2}\dfrac{2}{\left(x-2\right)\left(\sqrt{2x+5}+1\right)}=\dfrac{2}{\left(-2-2\right)\left(\sqrt[2]{2.\left(-2\right)+5}+1\right)}=\dfrac{2}{\left(-4\right).2}=-\dfrac{1}{4}\)
đk: \(x\le-2;x\ge0\)
Ta có: \(\sqrt{x\left(x+1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
\(\Leftrightarrow x\left(x+1\right)+2\sqrt{x\left(x+1\right)x\left(x+2\right)}+x\left(x+2\right)=4x^2\)
\(\Leftrightarrow x^2+x+2\sqrt{x^2\left(x^2+3x+2\right)}+x^2+2x=4x^2\)
\(\Leftrightarrow2x^2-3x=2\sqrt{x^4+3x^3+2x^2}\)
\(\Leftrightarrow\left(2x^2-3x\right)^2=4\left(x^4+3x^3+2x^2\right)\)
\(\Leftrightarrow4x^4-12x^3+9x^2=4x^4+12x^3+8x^2\)
\(\Leftrightarrow24x^3-x^2=0\)
\(\Leftrightarrow x^2\left(24x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\24x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=\frac{1}{24}\left(ktm\right)\end{cases}}\)
Vậy x = 0