(2x-3)^2022=(2x-3)^2021 tui cần tối hôm nay á sos
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VH
2
17 tháng 10 2021
\(M=2021+\left(x-2022\right)^{2022}\ge2021\forall x\)
Dấu '=' xảy ra khi x=2022
1 tháng 10 2021
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 7 2023
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 7 2023
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
NT
1
HN
0
NA
9 tháng 5 2022
`2x-15=-25`
`2x=-10`
`x=-5`
___________
`3/5<x/10<4/5`
`3/5=(3xx10)/(5xx10)=30/50`
`x/10=(5x)/(10xx5)=(5x)/50`
`4/5=(4xx10)/(5xx10)=40/50`
`=>30/50<(5x)/50<40/50`
`=>30<5x<40`
`=>x=7`
NH
22 tháng 2 2023
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 2 2023
A = \(\dfrac{2^{2021}+3^{2021}}{2^{2022}+3^{2022}}\)
Gọi ước chung lớn nhất của
22021 + 32021 và 22022+32022 là d (d\(\in\)N*)
Ta có : \(\left\{{}\begin{matrix}2^{2021}+3^{2021}⋮d\\2^{2022}+3^{2022}⋮d\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}2.(2^{2021}+3^{2021})⋮d\\2^{2022}+3^{2022}⋮d\end{matrix}\right.\)
Trừ vế với vế ta được 32022 - 2.32021 ⋮ d
⇒ 32021.( 3 - 2) ⋮ d
⇒ 32021 ⋮ d
⇒ d \(\in\){ 1; 3; 32; 33;........32021)
nếu d \(\in\) { 3; 32; 33;.....32021) thì
⇒ 22021 + 32021 ⋮ 3 ⇒ 22021 ⋮ 3 ( vô lý )
vậy d = 1
Hay phân số A = \(\dfrac{2^{2021}+3^{2021}}{2^{2022}+3^{2022}}\) là phân số tối giản (đpcm)
8 tháng 8 2023
a. \(\dfrac{2021+2020.2022}{2021.2022-1}\)
\(\dfrac{2021.2022-2022+2021}{2021.2022-1}=\dfrac{2021.2022-1}{2021.2022-1}=1\)
\(b.\dfrac{2022+2021.2023}{2022.2023-1}=\dfrac{2021.2023-2023+2022}{2022.2023-1}\)
\(=\dfrac{2021.2023-1}{2022.2023-1}\)
\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)
\(\left(2x-3\right)^{2021}\left(2x-3-1\right)=0\)
\(\left(2x-3\right)^{2021}\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
Ta có:
\(\left(2x-3\right)^{2022}=\left(2x-3\right)^{2021}\)
\(\Rightarrow\left(2x-3\right)^{2022}-\left(2x-3\right)^{2021}=0\)
\(\Rightarrow\left(2x-3\right)^{2021}\left[\left(2x-3\right)-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x-3-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{3}{2};2\right\}\).
\(Toru\)