Tìm x ∈ N, biết : (2x – 1 )^(2020) = ( 2x – 1 )^(2021) 29/11/2021 Bởi Valerie Tìm x ∈ N, biết : (2x – 1 )^(2020) = ( 2x – 1 )^(2021)
Đáp án: `x=1` Giải thích các bước giải: `(2x – 1 )^(2020) = ( 2x – 1 )^(2021)``=>(2x-1)^2020-(2x-1)^2021=0``=>(2x-1)^2020 – (2x-1)^2020 . (2x-1)=0``=>(2x-1)^2020 . 1-(2x-1)^2020 .(2x-1)=0``=>(2x-1)^2020.[1-(2x-1)]=0``=>`\(\left[ \begin{array}{l}(2x-1)^{2020}=0\\1-(2x-1)=0\end{array} \right.\)`=>`\(\left[ \begin{array}{l}2x-1=0\\2x-1=0+1\end{array} \right.\) `=>`\(\left[ \begin{array}{l}2x=0+1\\2x-1=1\end{array} \right.\) `=>`\(\left[ \begin{array}{l}2x=1\\2x=1+1\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1:2\\2x=2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=2:2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\) Vì `x\inNN` `=>x=1` Bình luận
Đáp án: `x=1` Giải thích các bước giải: Ta có : `(2x-1)^{2020}=(2x-1)^{2021}` `→(2x-1)^{2020}-(2x-1)^{2021}=0` `→(2x-1)^{2020}[1-(2x-1)]=0` `→(2x-1)^{2020}(1-2x+1)=0` `→(2x-1)^{2020}(2-2x)=0` `→` \(\left[ \begin{array}{l}(2x-1)^{2020}=0\\2-2x=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}2x-1=0\\2(1-x)=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}2x=1\\1-x=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=1\end{array} \right.\) Mà `x∈N` `→x=1` Bình luận
Đáp án:
`x=1`
Giải thích các bước giải:
`(2x – 1 )^(2020) = ( 2x – 1 )^(2021)`
`=>(2x-1)^2020-(2x-1)^2021=0`
`=>(2x-1)^2020 – (2x-1)^2020 . (2x-1)=0`
`=>(2x-1)^2020 . 1-(2x-1)^2020 .(2x-1)=0`
`=>(2x-1)^2020.[1-(2x-1)]=0`
`=>`\(\left[ \begin{array}{l}(2x-1)^{2020}=0\\1-(2x-1)=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}2x-1=0\\2x-1=0+1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}2x=0+1\\2x-1=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}2x=1\\2x=1+1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=1:2\\2x=2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=2:2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
Vì `x\inNN` `=>x=1`
Đáp án:
`x=1`
Giải thích các bước giải:
Ta có :
`(2x-1)^{2020}=(2x-1)^{2021}`
`→(2x-1)^{2020}-(2x-1)^{2021}=0`
`→(2x-1)^{2020}[1-(2x-1)]=0`
`→(2x-1)^{2020}(1-2x+1)=0`
`→(2x-1)^{2020}(2-2x)=0`
`→` \(\left[ \begin{array}{l}(2x-1)^{2020}=0\\2-2x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x-1=0\\2(1-x)=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x=1\\1-x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=1\end{array} \right.\)
Mà `x∈N`
`→x=1`