1 -1/(X+2)=2019/2020
Những câu hỏi liên quan
Nếu 1/3 + 1/6 +1/10 + ...... + 1/x.(x+1) : 2 = 2019/2021
A.x = 2019/2020 B. x = 2019 C. x = 2020 D. x = 2021
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a) 2019/2020 x 4/11 + 2019/2020 x 5/11 + 2019/2020 x 2/11
b) 17/14 x 25/27 - 1/14 x 25/27 - 2/14 x 25/27
\(a.=\dfrac{2019}{2020}\times\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)\\ =\dfrac{2019}{2020}\times1=\dfrac{2019}{2020}\\ b.=\dfrac{25}{27}\times\left(\dfrac{17}{14}-\dfrac{1}{14}-\dfrac{2}{14}\right)\\ =\dfrac{25}{27}\times1=\dfrac{25}{27}\)
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Tìm x,y biết x^2018+y^2018=x^2019+y^2019=x^2020+y^2020.
Cho a+b+c=2019, 1/a + 1/b+1/c=1/2019. Tính 1/a^2019+1/b^2019+1/c^2019
Tìm x,y biết x^2-xy=6x-5y-8.
Giúp mk với, mk vã lắm rồi :-( :-(
gt⇒x2−xy−(5x−5y)−x+8=0⇒(x−y)(x−5)−(x−5)=−3⇒(5−x)(x−y−1)=3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml"> là sẽ tìm được nghiệm nguyên của
2019 x 2020 - 1/ 2019 x 2020 và 2020 x 2021 - 1/ 2020 x 2021
so sánh phân số
8 tháng 10 2023 lúc 12:15
A=[ 2020 x 2019 + 2019 x 2018] x [ 1 + 1/2 : 1 và 1/2 - 1 và 1/3]
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
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Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Tìm x biết:
( 1/2 + 1/3 + ... + 1/2021 ).x = 2021/1 +2019/2 + ... + 2/2019 + 1/2020
Cho hàm số ydfrac{1}{3x^2-x-2}. Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?A. dfrac{2019!}{5}left(dfrac{1}{left(x-1right)^{2020}}-dfrac{3}{left(3x+2right)^{2020}}right)B. dfrac{2019!}{5}left(dfrac{3^{2020}}{left(3x+2right)^{2020}}-dfrac{1}{left(x-1right)^{2020}}right)C. dfrac{2019!}{5}left(dfrac{3}{left(3x+2right)^{2020}}-dfrac{1}{left(x-1right)^{2020}}right)D. dfrac{2019!}{5}left(dfrac{1}{left(x-1right)^{2020}}-dfrac{3^{2020}}{left(3x+2right)^{2020}}right)
Đọc tiếp
Cho hàm số \(y=\dfrac{1}{3x^2-x-2}\). Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?
A. \(\dfrac{2019!}{5}\left(\dfrac{1}{\left(x-1\right)^{2020}}-\dfrac{3}{\left(3x+2\right)^{2020}}\right)\)
B. \(\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
C. \(\dfrac{2019!}{5}\left(\dfrac{3}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
D. \(\dfrac{2019!}{5}\left(\dfrac{1}{\left(x-1\right)^{2020}}-\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}\right)\)
\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
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A=2020/20192+1 + 2020/20192+2 + 2020/20192+3 + ... + 2020/20192+2019. CMR 1<A<2.
Mục tiêu -500 sp mong giúp đỡ
So sánh x = 20192020 + 1 / 20192019 + 1 và y = 20192019 + 2020 / 20192018 + 2020
\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)
\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x>y\)