x+4/2020 +x+3/2020=x+2/2020+x+1/2023
tìm x
(x+1)/2023 + (x+2)/2022=(x+3)/2021 + (x+4)/2020
\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)
`(x+1)/2023+(x+2)/2022=(x+3)/2021+(x+4)/2020`
`=>(x+1)/2023+1+(x+2)/2022+1=(x+3)/2021+1+(x+4)/2020+1`
`=>(x+2024)/2023+(x+2024)/2022=(x+2024)/2021+(x+2024)/2020`
`=>(x+2024)/2023+(x+2024)/2022-(x+2024)/2021-(x+2024)/2020=0`
`=>(x+2024).(1/2023+1/2022-1/2021-1/2020)=0`
Vì `1/2023+1/2022-1/2021-1/2020` `\ne` `0`
`=> x+2024=0`
`=>x=-2024`
Giải phương trình sau: \(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
`<=>(x-1)/2023-1+(x-2)/2022-1=(x-3)/2021-1+(x-4)/2020-1`
`<=>(x-2024)/2023+(x-2024)/2022=(x-2024)/2021+(x-2024)/2020`
`<=>(x-2024)(1/2023+1/2022-1/2021-1/2020)=0`
`<=>x-2024=0(1/2023+1/2022-1/2021-1/2020>0)`
`<=>x=2024`
=>\(\left(\dfrac{x-1}{2023}-1\right)+\left(\dfrac{x-2}{2022}-1\right)=\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-4}{2020}-1\right)\)
=>x-2024=0
=>x=2024
\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
⇔\(\dfrac{x-1}{2023}-1+\dfrac{x-2}{2022}-1=\dfrac{x-3}{2021}-1+\dfrac{x-4}{2020}\)
⇔\(\dfrac{x-1}{2023}-\dfrac{2023}{2023}+\dfrac{x-2}{2022}-\dfrac{2022}{2022}=\dfrac{x-3}{2021}-\dfrac{2021}{2021}+\dfrac{x-4}{2020}-\dfrac{2020}{2020}\)
⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}=\dfrac{x-2024}{2021}+\dfrac{x-2024}{2020}\)
⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}-\dfrac{x-2024}{2021}-\dfrac{x-2024}{2020}=0\)
⇔\(\left(x-2024\right)\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\ne0\right)\)
⇔\(x-2024=0\)
⇔\(x=2024\)
Bài 4: tính giá trị biểu thức (1-1/2)x(1-1/3)x(1-1/4)x...x(1-1/2023) A.1 B.1/2023 C.1/2021 D.2020/2021
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
X+1/2020 + X+2/2019=X-1/2023 + X-2/2024
2022 x 2023 - 3 / 2023 x 2021 + 2020
cứu mình với
Lời giải:
\(\frac{2022\times 2023-3}{2023\times 2021+2020}=\frac{2023\times (2021+1)-3}{2023\times 2021+2020}
\\
=\frac{2023\times 2021+2023-3}{2023\times 2021+2020}=\frac{2023\times 2021+2020}{2023\times 2021+2020}=1\)
Tìm x, biết:
( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2023}\) ) . x = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) + \(\dfrac{2020}{3}\)
+ ... + \(\dfrac{1}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
1) Cho các số thực dương x,y,z thỏa mãn điều kiện x\(\sqrt{2020-y^2}\) + y\(\sqrt{2020-z^2}\) +z\(\sqrt{2023-x^2}\)=3030. Tính giá trị vủa biểu thức A=x\(^2\)+\(y^2\)+\(z^2\)
Lời giải:
Áp dụng BĐT AM-GM:
\(x\sqrt{2020-y^2}+y\sqrt{2020-z^2}+z\sqrt{2020-x^2}\leq \frac{x^2+(2020-y^2)}{2}+\frac{y^2+(2020-z^2)}{2}+\frac{z^2+(2020-x^2)}{2}=3030\)Dấu "=" xảy ra khi:
\(\left\{\begin{matrix} x^2=2020-y^2\\ y^2=2020-z^2\\ z^2=2020-x^2\end{matrix}\right.\Rightarrow x=y=z=\sqrt{1010}\)
Khi đó:
$A=3(\sqrt{1010})^2=3030$
tìm x nguyên 2023+2022+2021+2020+...+x=2023
Bài 1 :Tìm x , biết :
\(\dfrac{\left(2020^{100}+2020^{96}+2020^{92}+...+2020^4+1\right)}{\left|x-2020\right|}\) = \(\dfrac{2020^{104}-1}{2020^4-1}\)
Bài 2 : So sánh phân số 111979 và 371320
Bài 3 : Trong tập hợp số tự nhiên có thể số có dạng 20202020....20200....0 chia hết cho 2021 hay không ?
Bài 2:
Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)
\(37^{1320}=37^{2\cdot660}=1369^{660}\)
mà \(1331^{660}< 1369^{660}\)
nên \(11^{1979}< 37^{1320}\)