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\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)
\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)
\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)
\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
Đặt \(f\left(x\right)=\left(2m^2-5m+2\right)\left(x+2\right)^{2019}\left(x^{2020}-4\right)+5x-1\)
Do \(f\left(x\right)\) là hàm đa thức nên liên tục và xác định trên R
Ta có: \(f\left(-2\right)=-11< 0\)
Lại có: \(4>1\Rightarrow\sqrt[2020]{4}>1\Rightarrow5\sqrt[2020]{4}>1\)
\(\Rightarrow f\left(\sqrt[2020]{4}\right)=5\sqrt[2020]{4}-1>0\)
\(\Rightarrow f\left(-2\right).f\left(\sqrt[2020]{4}\right)< 0;\forall m\)
\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(-2;\sqrt[2020]{4}\right)\) hay pt đã cho luôn luôn có nghiệm với mọi m
\(\frac{1}{2}.cos\left(x+\frac{\pi}{5}\right)+\frac{\sqrt{3}}{2}.sin\left(x+\frac{\pi}{5}\right)=0\)
\(\Leftrightarrow cos\left(\frac{\pi}{3}\right).cos\left(x+\frac{\pi}{5}\right)+sin\left(\frac{\pi}{3}\right).sin\left(x+\frac{\pi}{5}\right)=0\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{5}-\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow cos\left(x-\frac{2\pi}{15}\right)=0\)
\(\Leftrightarrow x-\frac{2\pi}{15}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{19\pi}{30}+k\pi\) \(k\in\left[-2020;-1\right]\)
a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)
b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)
c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)
d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)
e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)
tổng chữ số hàng đơn vị và hàng chục của limx->1 \(\frac{x^{2020}-2020x+ 2019}{\left(x-1\right)^2}\)
\(\lim\limits_{x\rightarrow1}\frac{x^{2020}-2020x+2019}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{2020x^{2019}-2020}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{2019.2020.x^{2018}}{2}=\frac{2019.2020}{2}=2039190\)
Tổng chữ số hàng chục và đơn vị là 9
Lời giải:
\(\lim\limits_{x\to \pm\infty}\sqrt{x^2-3x+4}=\lim\limits_{x\to \pm\infty}\sqrt{x^2}.\lim\limits_{x\to \pm \infty}\sqrt{1-\frac{3}{x}+\frac{4}{x^2}}=\lim\limits_{x\to \pm\infty}|x|.1=+\infty \)
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\(\lim\limits_{x\to +\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to +\infty}x^2.\lim\limits_{x\to +\infty}(\sqrt{1+\frac{5}{x^2}}+1)=2(+\infty )=+\infty \)
\(\lim\limits_{x\to -\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to -\infty}\frac{5x}{\sqrt{x^2+5}-x}=\lim\limits_{x\to -\infty}\frac{-5}{\sqrt{1+\frac{5}{x^2}}+1}=\frac{-5}{2}\)
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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
Lời giải:
\(\lim\limits_{x\to \pm\infty}\sqrt{x^2-3x+4}=\lim\limits_{x\to \pm\infty}\sqrt{x^2}.\lim\limits_{x\to \pm \infty}\sqrt{1-\frac{3}{x}+\frac{4}{x^2}}=\lim\limits_{x\to \pm\infty}|x|.1=+\infty \)
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\(\lim\limits_{x\to +\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to +\infty}x^2.\lim\limits_{x\to +\infty}(\sqrt{1+\frac{5}{x^2}}+1)=2(+\infty )=+\infty \)
\(\lim\limits_{x\to -\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to -\infty}\frac{5x}{\sqrt{x^2+5}-x}=\lim\limits_{x\to -\infty}\frac{-5}{\sqrt{1+\frac{5}{x^2}}+1}=\frac{-5}{2}\)
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\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
\(S=(C_{2020}^1)^2+(2C_{2020}^2)^2+(3C_{2020}^3)^2+\ldots+(2020C_{2020}^{2020})^2\)
Ta có đẳng thức tổ hợp sau:
\(k C_n^k = k \cdot \frac{n!}{k!(n-k)!} = n \cdot \frac{(n-1)!}{(k-1)!(n-k)!} = n C_{n-1}^{k-1}\)
=>\(S = \sum_{k=1}^{n} (k C_n^k)^2 = \sum_{k=1}^{n} (n C_{n-1}^{k-1})^2 = n^2 \sum_{k=1}^{n} \left( C_{n-1}^{k-1} \right)^2\)
Đặt i=k-1
Khi k chạy từ 1 đến n
thì i chạy từ 0 đến n-1
=>\(S = n^2 \sum_{i=0}^{n-1} \left( C_{n-1}^i \right)^2\)
Đặt \(T = \sum_{i=0}^{n-1} \left( C_{n-1}^i \right)^2\)
\((1+x)^{n-1} = C_{n-1}^0 + C_{n-1}^1 x + C_{n-1}^2 x^2 + \ldots + C_{n-1}^{n-1} x^{n-1}\)
\((x+1)^{n-1} = C_{n-1}^0 x^{n-1} + C_{n-1}^1 x^{n-2} + \ldots + C_{n-1}^{n-1}\)
Do đó: \(\left(1+x\right)^{2n-2}=(C_{n-1}^0)^2+(C_{n-1}^1)^2+(C_{n-1}^2)^2+\ldots+(C_{n-1}^{n-1})^2=\sum_{i=0}^{n-1}\left(C_{n-1}^{i}\right)^2\)
\(=\sum_{i=0}^{n-1}\left(C_{n-1}^{i}\right)^2=C_{2n-2}^{n-1}\)
Do đó: \(S = n^2 C_{2n-2}^{n-1}\)
\(=2020^2C_{2(2020)-2}^{2020-1}=2020^2C_{4038}^{2019}\)