Tính nhanh :
a,2017 x 2021 - 4031 / 2020 + 2017 x 2018
b,2017 x 2019 + 1009 / 2019 x 4035 - 1
Những câu hỏi liên quan
Tính nhanh :
a,2017 x 2021 - 4031 / 2020 + 2017 x 2018
b,2017 x 2019 + 1009 / 2019 x 4035 - 1
a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)
= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)
= 1
@Nguyen Thi Ngoc Linh
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x-2017/2019 + x-2019/2017 = x+6/2021
x-2017/2019 + x-2019/2017 = x+6/2021
\(\dfrac{x-2017}{2019}+\dfrac{x-2019}{2017}=\dfrac{x+6}{2021}\)
\(\Rightarrow\dfrac{x-2017}{2019}-1+\dfrac{x-2019}{2017}-1=\dfrac{x+6}{2021}-2\)
\(\Rightarrow\dfrac{x-2017}{2019}-\dfrac{2019}{2019}+\dfrac{x-2019}{2017}-\dfrac{2017}{2017}=\dfrac{x+6}{2021}-\dfrac{4042}{2021}\)
\(\Rightarrow\dfrac{x-2017-2019}{2019}+\dfrac{x-2019-2017}{2017}=\dfrac{x+6-4042}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2019}+\dfrac{x-4036}{2017}=\dfrac{x-4036}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2021}-\dfrac{x-4036}{2019}-\dfrac{x-4036}{2017}=0\)
\(\Rightarrow\left(x-4036\right)\left(\dfrac{1}{2021}-\dfrac{1}{2019}-\dfrac{1}{2017}\right)=0\)
=> x - 4036 = 0
=> x = 4036
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x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017 − 2019/ 2019 + x − 2019 − 2017/2017
= x + 6 − 4042/2021
=> x − 2017 − 2019/2019 + x − 2019 − 2017/2017 = x + 6 − 4042/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=>(x − 4036)(12021 − 12019 − 12017) = 0
=> x - 4036 = 0
=> x = 4036
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cho x,y,z thỏa mãn : x+y+z=1/2 ; 1/y^2+1/z^2+1/xyz=4 ; 1/x+1/y+1/z>0. tính Q = (x^2019+z^2019)+(y^2017+z^2017)(x^2021+y^2021)
Tìm GTNN của biểu thức: P=|x-2017|+|x-2019|+|x-2020|+|y-2021|
Hình như là vậy á
Chúc bạn học tốt
B=x^2020 -2019 x^2019 - x^2018 - 2019 x^2017 - ...-2019x-2020 với x=2020
Cho 3 số x, y, z TM: \(\left\{{}\begin{matrix}x+y+z=2017\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2017}\end{matrix}\right.\)
Tính GTBT: \(P=\left(x^{2017}+y^{2017}\right)\left(y^{2019}+z^{2019}\right)\left(z^{2021}+x^{2021}\right)\)
Cho 3 số x,y,z thỏa mãn \(\hept{\begin{cases}x+y+z=2019\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2019}\end{cases}}\).Tính giá trị biểu thức \(P=\left(x^{2017}+y^{2017}\right)\left(y^{2019}+z^{2019}\right)\left(z^{2021}+x^{2021}\right)\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z}{\left(x+y+z\right).z}-\frac{x+y+z}{z.\left(x+y+z\right)}=\frac{-x-y}{z.\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{x+y}{-z.\left(x+y+z\right)}\)
TH1: x+y=0
=> x=-y => P=0
TH2: xy=-z.(x+y+z)
\(\Leftrightarrow xy=-xz-zy-z^2\Leftrightarrow xy+xz+zy+z^2=0\Leftrightarrow x.\left(y+z\right)+z.\left(y+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(y+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-z\\y=-z\end{cases}\Rightarrow P=0}\)
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cho p(x)=x^2020-2019x^2019+x^2018-2019x^2017+...+x^2019 tinh P(2019)