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\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\\ \Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
`#3107.101107`
\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\)
\(\Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\cdot\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
Vậy, \(x\in\left\{4;5;6\right\}.\)
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
Ta có tính chất: \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)
\(A=\dfrac{2022^{99}-1}{2022^{100}-1}>\dfrac{2022^{99}-1-2021}{2022^{100}-1-2021}\)
\(A>\dfrac{2022^{99}-2022}{2022^{100}-2022}\)
\(A>\dfrac{2022\left(2022^{98}-1\right)}{2022\left(2022^{99}-1\right)}\)
\(A>\dfrac{2022^{98}-1}{2022^{99}-1}\)
\(A>B\)
\(S=1+3^2+3^4+...+3^{2022}\)
\(3^2S=9S=3^2+3^4+3^6+...+3^{2024}\)
\(S=\dfrac{9S-S}{8}=\left(3^{2024}-1\right):8\)
d, không đáp án nào đúng
Lời giải:
$S=1+3^2+3^4+....+3^{2022}$
$9S=3^2S=3^2+3^4+3^6+...+3^{2024}$
$\Rightarrow 9S-S=3^{2024}-1$
$\Rightarrow S=\frac{3^{2024}-1}{8}$
Đáp án D.
a)\(27.65+27.35+300=27.\left(65+35\right)+300\)
\(=27.100+300=2700+300=3000\)
b)\(3838:\left[\left(190-6.5^2\right):4+3\right]\)
\(=3838:\left[\left(190-6.25\right):4+3\right]\)
\(=3838:\left[\left(190-150\right):4+3\right]\)
\(=3838:\left[40:4+3\right]=3838:\left[10+3\right]\)
\(=3838:13=\dfrac{3838}{13}\)
c)\(2022-x=2021\)
\(x=2022-2021=1\)
d)\(26+14:\left(x-5\right)=33\)
\(14:\left(x-5\right)=33-26=7\)
\(x-5=14+7=2\)
\(x=2+5=7\)
e)đề hỏi làm j thế bạn
\(=\dfrac{-2}{3}\cdot\dfrac{12}{19}-\dfrac{2}{3}\cdot\dfrac{7}{19}+1\)
=-2/3(12/19+7/19)+1
=1-2/3
=1/3
\(a,TH1:x-2021=0=>x=2021\)
\(Th2:x-2022=0=>x=2022\)
Vậy \(x\in\left\{2021;2022\right\}\)
\(b,x\left(8-5\right)=1080\)
\(x.3=1080\)
\(x=360\)
\(c,x^3=216< =>6^3=216=>x=3\)
\(d,5^5=3125\)
a) ( x- 2021) * ( x- 2022) = 0
=> \(\orbr{\begin{cases}x-2021=0\\x-2022=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2021\\x=2022\end{cases}}}\)
b) b. 8x - 5x = 2022
=> 3x = 2022
=> x = 674
c) \(5\cdot x^3=1080\)
=> \(x^3=216\)
=> \(x^3=6^3\)
=> x = 6
d) \(5^x=3125\)
=> \(5^x=5^5\)
=> x = 5
x2022 = x
⇔ x2022 - x = 0
⇔ x.(x2021 - 1) = 0
⇔ x=0; hoặc x2021 = 1
⇔ x = 0; x = 1
Lời giảiL
$x^{2022}=x$
$x^{2022}-x=0$
$x(x^{2021}-1)=0$
$\Rightarrow x=0$ hoặc $x^{2021}=1$
Hay $x=0$ hoặc $x=1$