Tính nhanh A =2021 phần 2022 x 7 phần 8 + 2021 phần 2022 x 3 phần 8 - 2021 phần 2022 x 1 phần 4
so sánh 2022 mũ 2023 +1 phần 2022 mũ 2021 +1 Với 2022 mũ 2021 +1 phần 2022 mũ 2019 +1.help me, giúp với ,khó quá ko làm đc
Tính A=1-2-3+4-5-6+7-8-9+...+2022-2021-2022
A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022
A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)
A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)
A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]
A=674.(−1013,5)D=674.(-1013,5)
A=−683099
A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022
A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)
A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)
A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]
A=674.(−1013,5)D=674.(-1013,5)
A=−683099
Tính nhanh:
2022 x 2023 - 1/2023 x 2021 + 2022
\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)
= 1
tính nhanh: ( 2022 x 0,75 + 2022 : 4) - ( 2021 : 0,1 : 10}
A = ( 2022 x 0,75 + 2022 : 4) - ( 2021 : 0,1 : 10)
= (2022 x 0,75 + 2022 x 0,25) - ( 2021 : 1/10 : 10)
= 2022 x(0,75+0,25) - 2021x 10 : 10
= 2022x 1 - 2021
= 2022 - 2021
=1
Tính nhanh: (2022 x 2021 – 2021 x 2020) x( 1 + \(\dfrac{1}{2}\) : \(1\dfrac{1}{2}\) - \(1\dfrac{1}{3}\) )
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
\(\dfrac{-6}{17}x\dfrac{-2021}{2022}+\dfrac{2021}{2022}x\dfrac{-23}{17}+\dfrac{2021}{2022}\)
\(=\dfrac{2021}{2022}\left(\dfrac{6}{17}-\dfrac{23}{17}\right)+\dfrac{2021}{2022}=\dfrac{-2021}{2022}+\dfrac{2021}{2022}=0\)
tìm x: 3(5x+8)^4+2022=2021
ai giải giúp mình đi mình đang cần gấp
Ta có: \(3\left(5x+8\right)^4+2022=2021\)
\(\Leftrightarrow3\left(5x+8\right)^4=-1\)(Vô lý)
Vậy: \(x\in\varnothing\)
x+1/2021*2022+1/2021*2022+......+1/3*2+1/3*2=1
cho x,y,z khác 0 thoả mãn x+y+z=2022 và 1/x+1/y+1/z=1/2022 CMR: 1/x^2021+1/y^2021+1/z^2021=1/x^2021+y^2021+z^2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2022}\)
\(\Rightarrow\dfrac{yz+zx+xy}{xyz}=\dfrac{1}{x+y+z}\)
\(\Rightarrow\left(yz+zx+xy\right)\left(x+y+z\right)=xyz\)
\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+3xyz-xyz=0\)
\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Rightarrow x=-y\) hoặc \(y=-z\) hoặc \(z=-x\).
-Đến đây thôi bạn, câu hỏi sai rồi ạ.
tìm x, y thuộc Z biết (x-2021)^2+(x-2022)^2022=2022^y-2021