\(\left|x-1\right|^{2019}+\left|x\right|^{2020}=-1\)
tìm x biết
\(\frac{\left(2019-x^2\right)+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020^2\right)}\) = \(\frac{19}{49}\)
Tìm x biết \(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)}\)\(\frac{+\left(x-2020\right)^2}{+\left(x-2020\right)^2}\)\(=\frac{19}{49}\)
Ta có: \(f\left(2019\right)=2020=2019+1\)
\(f\left(2020\right)=2021=2020+1\)
Đặt \(h\left(x\right)=-x-1\)và \(g\left(x\right)=f\left(x\right)+h\left(x\right)\)
\(\Rightarrow\hept{\begin{cases}g\left(2019\right)=f\left(2019\right)+h\left(2019\right)=2020-2020=0\\g\left(2020\right)=f\left(2020\right)+h\left(2020\right)=2021-2021=0\end{cases}}\)
\(\Rightarrow x=2019;x=2020\)là nghiệm của đa thức g(x) mà g(x) là đa thức bậc 3 , hệ số \(x^3\)là số nguyên
\(\Rightarrow g\left(x\right)=a\left(x-2019\right)\left(x-2020\right)\left(x-x_0\right)\)(\(a\in\)Z*)
\(\Rightarrow f\left(x\right)=g\left(x\right)-h\left(x\right)\)
\(=a\left(x-2019\right)\left(x-2020\right)\left(x-x_0\right)+x+1\)
\(f\left(2021\right)=a\left(2021-2019\right)\left(2021-2020\right)\left(2021-x_0\right)+2021+1\)
\(=a.1.2\left(2021-x_0\right)+2022\)
\(f\left(2018\right)=a\left(2018-2019\right)\left(2018-2020\right)\left(2018-x_0\right)+2018+1\)
\(=a.1.2.\left(2018-x_0\right)+2019\)
\(\Rightarrow f\left(2021\right)-f\left(2018\right)=a.1.2\left(2021-2018\right)+3\)
\(=6a+3\)
Làm nốt
Cho đa thức \(f\left(x\right)\)bậc 3 với hệ số \(x^3\)là số nguyên dương thỏa mãn:
\(f\left(2019\right)=2020;f\left(2020\right)=2021\)
CMR \(f\left(2021\right)-f\left(2018\right)\)là hợp số
Tìm x, biết:
\(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}=\frac{19}{49}\)
Các bạn mong giúp mình sớm nhé
ủa bạn j ơi chữ x chành bành ra trên đề kìa mà bạn bảo tìm làm j nữa
Tham khảo tại: Câu hỏi của Lương Đức Hưng - Toán lớp 8 | Học trực tuyến
Cho hàm số \(y=\dfrac{1}{2x^2+x-1}\). Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?
A. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2019}}{\left(2x-1\right)^{2020}}\right)\)
B. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
C. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2}{\left(2x-1\right)^{2020}}\right)\)
D. \(\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}+\dfrac{2}{\left(2x-1\right)^{2020}}\right)\)
\(y=\dfrac{1}{2x^2+x-1}=\dfrac{1}{\left(x+1\right)\left(2x-1\right)}=\dfrac{2}{3}.\dfrac{1}{2x-1}-\dfrac{1}{3}.\dfrac{1}{x+1}\)
\(y'=\dfrac{2}{3}.\dfrac{-2}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{-1}{\left(x+1\right)^2}=\dfrac{2}{3}.\dfrac{\left(-1\right)^1.2^1.1!}{\left(2x-1\right)^2}-\dfrac{1}{3}.\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}\)
\(y''=\dfrac{2}{3}.\dfrac{\left(-1\right)^2.2^2.2!}{\left(2x-1\right)^3}-\dfrac{1}{3}.\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^n.2^n.n!}{\left(2x-1\right)^{n+1}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{2}{3}.\dfrac{\left(-1\right)^{2019}.2^{2019}.2019!}{\left(2x-1\right)^{2020}}-\dfrac{1}{3}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}\)
\(=\dfrac{2019!}{3}\left(\dfrac{1}{\left(x+1\right)^{2020}}-\dfrac{2^{2020}}{\left(2x-1\right)^{2020}}\right)\)
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. \(\dfrac{2019!}{5}\left(\dfrac{1}{\left(x-1\right)^{2020}}-\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}\right)\)
\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)
\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)
\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)
\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
1,tìm x biết:
\(\left|x-2019\right|^{2019}+\left|x-2020\right|^{2020}=1\)
Ta có: |x - 2019| ≥ 0 => |x - 2019|2019 ≥ 0
|x - 2020| ≥ 0 => |x - 2020|2020 ≥ 0
+) TH1: \(\hept{\begin{cases}\left|x-2019\right|^{2019}=0\\\left|x-2020\right|^{2020}=1\end{cases}\Rightarrow}\hept{\begin{cases}\left|x-2019\right|=0\\\left|x-2020\right|=1\end{cases}}\Rightarrow\hept{\begin{cases}x-2019=0\\\left|x-2020\right|=1\end{cases}\Rightarrow}\hept{\begin{cases}x=2019\\\left|x-2020\right|=1\end{cases}}\)
Giải: |x - 2020| = 1
TH1: x - 2020 = 1 => x = 2021
TH2: x - 2020 = -1 => x = 2019
Vì 2021 ≠ 2019
=> x = 2019
+) TH2: \(\hept{\begin{cases}\left|x-2019\right|^{2019}=1\\\left|x-2020\right|^{2020}=0\end{cases}\Rightarrow}\hept{\begin{cases}\left|x-2019\right|=1\\\left|x-2020\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-2019\right|=1\\x-2020=0\end{cases}\Rightarrow}\hept{\begin{cases}\left|x-2019\right|=1\\x=2020\end{cases}}\)
Giải |x - 2019| = 1
Th1: x - 2019 = 1 => x = 2020
Th2: x - 2019 = -1 => x = 2018
Vì 2018 ≠ 2020
=> x = 2020
Vậy x \(\in\){ 2020; 2019 }
P/s: Ko chắc :)
Trả lời :
Bạn Kan làm đúng rồi nha !
Học tốt
#Sơn%#
Trả lời
Bạn làm theo bài bạn Kan nha
bạn ý làm đúng rồi
hok tốt
Tính
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
( ĐKXĐ : \(x\ne\left\{0;-1;-2;...;-2019;-2020\right\}\))
\(=\frac{1}{x}-\frac{1}{\left(x+1\right)}+\frac{1}{\left(x+1\right)}-\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)}-\frac{1}{\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+2020}\)
\(=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x\left(x+2020\right)}\)
\(=\frac{x+2020-x}{x\left(x+2020\right)}\)
\(=\frac{2020}{x\left(x+2020\right)}\)
Bài giải
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2019}-\frac{1}{x+2020}\)
\(=\frac{1}{x}-\frac{1}{x+2020}\)
\(=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x+2020}=\frac{2020}{x\left(x+2020\right)}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+2019\right)\left(x+2020\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+2019}-\frac{1}{x+2020}\)
\(=\frac{1}{x}-\frac{1}{x+2020}=\frac{x+2020}{x\left(x+2020\right)}-\frac{x}{x\left(x+2020\right)}=\frac{2020}{x\left(x+2020\right)}\)
1. Giải phương trình nghiệm nguyên
a) \(x^2+4x+2018^{10}\)
b) \(x^2+4x+\left(y-1\right)^2=21\)
c) \(x^2+3\left(y-1\right)^2=2021\)
d) \(\left(3x-1\right)^{2020}-18\left(y-2\right)^{2019}=2019^{2020}\)
2. Tìm x,y ∈ Z
a) \(x^2-y^2+6y=56\)
b) \(x^2-4x+9y^2-6y=11\)
\(1,\\ b,\Leftrightarrow\left(x^2+4x+4\right)+\left(y-1\right)^2=25\\ \Leftrightarrow\left(x+2\right)^2+\left(y-1\right)^2=25\)
Vậy pt vô nghiệm do 25 ko phải tổng 2 số chính phương
\(2,\\ a,\Leftrightarrow x^2-\left(y^2-6y+9\right)=47\\ \Leftrightarrow x^2-\left(y-3\right)^2=47\)
Mà 47 ko phải hiệu 2 số chính phương nên pt vô nghiệm
\(b,\Leftrightarrow\left(x-2\right)^2+\left(3y-1\right)^2=16\)
Mà 16 ko phải tổng 2 số chính phương nên pt vô nghiệm
1a. Đề lỗi
1b.
PT $\Leftrightarrow (x+2)^2+(y-1)^2=25$
$\Leftrightarrow (x+2)^2=25-(y-1)^2\leq 25$
$(x+2)^2$ là scp không vượt quá $25$ nên có thể nhận các giá trị $0,1,4,9,16,25$
Nếu $(x+2)^2=0\Rightarrow (y-1)^2=25$
$\Rightarrow (x,y)=(-2, 6), (-2, -4)$
Nếu $(x+2)^2=1\Rightarrow (y-1)^2=24$ không là scp (loại)
Nếu $(x+2)^2=4\Rightarrow (y-1)^2=21$ không là scp (loại)
Nếu $(x+2)^2=9\Rightarrow (y-1)^2=16$
$\Rightarrow (x,y)=(1, 5), (1, -3), (-5,5), (-5, -3)$
Nếu $(x+2)^2=25\Rightarrow (y-1)^2=0$
$\Rightarrow (x,y)=(3, 1), (-7, 1)$
1c.
Vì $x^2$ là scp nên $x^2\equiv 0,1\pmod 3$
$3(y-1)^2\equiv 0\pmod 3$
$\Rightarrow x^2+3(y-1)^2\equiv 0,1\pmod 3$
Mà $2021\equiv 2\pmod 3$
Do đó pt $x^2+3(y-1)^2=2021$ vô nghiệm
1d.
Ta thấy:
$(3x-1)^{2020}$ là scp không chia hết cho $3$ nên $(3x-1)^{2020}\equiv 1\pmod 3$
$18(y-2)^{2019}\equiv 0\pmod 3$
$\Rightarrow (3x-1)^{2020}+18(y-2)^{2019}\equiv 1\pmod 3$
Mà $2019^{2020}\equiv 0\pmod 3$
Do đó pt vô nghiệm.
\(\left(x-2020\right)^{x-1}-\left(x-2020\right)^{x+2019}=0\)0
(x-2020)x - 1 - (x - 2020)x + 2019 = 0
=> (x - 2020)x - 1 .[(x - 2020)2020 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-2020\right)^{x-1}=0\\\left(x-2020\right)^{2020}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2020=0\\\left(x-2020\right)^{2020}=1^{2020}\end{cases}\Rightarrow}\orbr{\begin{cases}x-2020=0\\x-2020=\pm1\end{cases}}}\)
=> \(x-2020\in\left\{0;1;-1\right\}\Rightarrow x\in\left\{2020;2021;2019\right\}\)