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\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)
\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)
\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)
\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
Ta có: \(u_n>2020\) với mọi \(n\in N\text{*}\) \(\left(\text{*}\right)\)
Thật vậy, dễ thấy \(u_1=2021>2020\)
Giả sử \(\left(\text{*}\right)\) đúng với \(n=k\left(k\ge1\right)\)
\(\Rightarrow u_k>2020\)\(\Rightarrow u_{k+1}=\left[1-\dfrac{1}{\left(k+1\right)^2}\right]u_k+\dfrac{2020}{\left(k+1\right)^2}\)
\(>\left[1-\dfrac{1}{\left(k+1\right)^2}\right].2020+\dfrac{2020}{\left(k+1\right)^2}=2020\)
\(\Rightarrow\left(\text{*}\right)\) đúng với \(n=k+1\)
Do đó theo nguyên lý quy nạp ta có đpcm.
Lại có:
\(u_{n+1}-u_n=\dfrac{2020}{\left(n+1\right)^2}-\dfrac{u_n}{\left(n+1\right)^2}< 0\) với mọi \(n\in N\text{*}\)
\(\Rightarrow\left(u_n\right)\) là dãy giảm
\(\left(u_n\right)\) là dãy giảm và bị chặn nên \(\left(u_n\right)\) là dãy hội tụ
Đặt \(limu_n=L\)
\(\Rightarrow\left\{{}\begin{matrix}2020\le L\le2021\\L=\left[1-\dfrac{1}{\left(n+1\right)^2}\right].L+\dfrac{2020}{\left(n+1\right)^2}\end{matrix}\right.\)\(\Rightarrow L=2020\left(tm\right)\)
Vậy \(limu_n=2020\)
Ta có: \(u_n>2020\) với mọi \(n\in N\text{*}\) \(\left(\text{*}\right)\)
Thật vậy, dễ thấy \(u_1=2021>2020\)
Giả sử \(\left(\text{*}\right)\) đúng với \(n=k\left(k\ge1\right)\)
\(\Rightarrow u_k>2020\)\(\Rightarrow u_{k+1}=\left[1-\dfrac{1}{\left(k+1\right)^2}\right]u_k+\dfrac{2020}{\left(k+1\right)^2}\)
\(>\left[1-\dfrac{1}{\left(k+1\right)^2}\right].2020+\dfrac{2020}{\left(k+1\right)^2}=2020\)
\(\Rightarrow\left(\text{*}\right)\) đúng với \(n=k+1\)
Do đó theo nguyên lý quy nạp ta có đpcm.
Lại có:
\(u_{n+1}-u_n=\dfrac{2020}{\left(n+1\right)^2}-\dfrac{u_n}{\left(n+1\right)^2}< 0\) với mọi \(n\in N\text{*}\)
\(\Rightarrow\left(u_n\right)\) là dãy giảm
\(\left(u_n\right)\) là dãy giảm và bị chặn nên \(\left(u_n\right)\) là dãy hội tụ
Đặt \(limu_n=L\)
\(\Rightarrow\left\{{}\begin{matrix}2020\le L\le2021\\L=\left[1-\dfrac{1}{\left(n+1\right)^2}\right].L+\dfrac{2020}{\left(n+1\right)^2}\end{matrix}\right.\)\(\Rightarrow L=2020\left(tm\right)\)
Vậy \(limu_n=2020\)
a/ \(y=\left(x^3-3x\right)^{\dfrac{3}{2}}\Rightarrow y'=\dfrac{3}{2}\left(x^3-3x\right)^{\dfrac{1}{2}}\left(x^3-3x\right)'=\dfrac{3}{2}\left(3x^2-3\right)\sqrt{x^3-3x}\)
b/ \(y'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\sqrt{x^3+1}-x^2+2\right)'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\dfrac{3x^2}{\sqrt{x^3+1}}-2x\right)\)c/
\(y'=14\left(x^6+2x-3\right)^6\left(x^6+2x-3\right)'=14\left(x^6+2x-3\right)^6\left(6x^5+2\right)\)
d/ \(y=\left(x^3-1\right)^{-\dfrac{5}{2}}\Rightarrow y'=-\dfrac{5}{2}\left(x^3-1\right)^{-\dfrac{7}{2}}\left(x^3-1\right)'=-\dfrac{15x^2}{2\sqrt{\left(x^3-1\right)^7}}\)
a. Làm gọn 1 chút xíu:
\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)
\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)
b.
\(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)
c.
\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)