tìm giá trị lớn nhất của P = \(\dfrac{|x-2022|-|x-2023|+|x-2024|+2022}{|x-2022|+|x-2023|+|x-2024|}\)
\(\dfrac{-6}{17}x\dfrac{-2021}{2022}+\dfrac{2021}{2022}x\dfrac{-23}{17}+\dfrac{2021}{2022}\)
\(=\dfrac{2021}{2022}\left(\dfrac{6}{17}-\dfrac{23}{17}\right)+\dfrac{2021}{2022}=\dfrac{-2021}{2022}+\dfrac{2021}{2022}=0\)
Tính giá trị của biểu thức sau: B= \(\dfrac{tan\left(\dfrac{23\pi}{2}+x\right).sin\left(2022\pi-x\right).cos\left(x-2021\pi\right)}{cos\left(\dfrac{2021\pi}{2}-x\right).sin\left(x+2023\pi\right)}\)
\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)
\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)
\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
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
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
\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
⇔\(\dfrac{x-1}{2023}-1+\dfrac{x-2}{2022}-1=\dfrac{x-3}{2021}-1+\dfrac{x-4}{2020}\)
⇔\(\dfrac{x-1}{2023}-\dfrac{2023}{2023}+\dfrac{x-2}{2022}-\dfrac{2022}{2022}=\dfrac{x-3}{2021}-\dfrac{2021}{2021}+\dfrac{x-4}{2020}-\dfrac{2020}{2020}\)
⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}=\dfrac{x-2024}{2021}+\dfrac{x-2024}{2020}\)
⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}-\dfrac{x-2024}{2021}-\dfrac{x-2024}{2020}=0\)
⇔\(\left(x-2024\right)\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\ne0\right)\)
⇔\(x-2024=0\)
⇔\(x=2024\)
\(\dfrac{2022}{2023}\)+\(\dfrac{2023}{2024}\)+\(\dfrac{2024}{2022}\)
So sánh : \(A=\dfrac{8^{2021}+2}{8^{2022}+2}\) với \(B=\dfrac{8^{2023}+2}{8^{2024}+2}\)
Giúp với
\(8A=\dfrac{8^{2022}+16}{8^{2022}+2}=1+\dfrac{14}{8^{2022}+2}\)
\(8B=\dfrac{8^{2024}+16}{8^{2024}+2}=1+\dfrac{14}{8^{2024}+2}\)
Vì \(\dfrac{14}{8^{2022}+2}>\dfrac{14}{8^{2024}+2}\)
=> 8A>8B
=> A>B
\(\dfrac{x+1}{2022}\)+\(\dfrac{x+2}{2021}\)+.....+\(\dfrac{x+23}{2000}\)+23=0
Mọi người giúp mik vs ạ,mai mik phải nộp rồi ạ!
Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$
$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$
$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương
$\Rightarrow x+2023=0$
$\Leftrightarrow x=-2023$
Cho x,y,z khác 0 thỏa mãn x+yz=2022 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2022\)
CMR: \(\dfrac{1}{x^{2021}}+\dfrac{1}{y^{2021}}+\dfrac{1}{z^{2021}}=\dfrac{1}{x^{2021}+y^{2021}+z^{2021}}\)