\(\left(2x-1\right)^2+17=18\\ \Leftrightarrow\left(2x-1\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
![⌛メ𝕹𝖊𝖕𝖆𝖑 ツ[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]](https://hoc24.vn/images/avt/avt31903185_256by256.jpg)
\(\left(2x-1\right)^2+17=18\)
\(\Leftrightarrow\left(2x-1\right)^2-1=0\)
\(\Leftrightarrow\left(2x-1-1\right)\left(2x-1+1\right)=0\)
\(\Leftrightarrow2x\left(2x-2\right)=0\)
\(\Rightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
TL
\((2x - 1)^2\) +17 =18
\((2x-1)^2\) =18-17
\((2x-1)^2\) =1
\((2x-1)^2\) =\(1^2\)
2x-1 =1
2x =1+1
2x =2
x =2:2
x =1
nha bn
HT
đáp án
\(x=\frac{3}{4}\)
hok tốt nha
ok
nha
@@@@@
Kết quả =
3/4
Đúng không nên tíc cho san
TL:
x =3/4
HT
K mik nha
\(\left(2x-1\right)^2+17=18\)
\(\left(2x-1\right)^2=18-17\)
\(\left(2x-1\right)^2=1\)
Ta có: \(1=1^2=\left(-1\right)^2\)
\(\left(2x-1\right)^2=1\) \(\left(2x-1\right)^2=1^2\) \(\left(2x-1\right)=1\) \(\Rightarrow2x-1=1\) \(2x=1+1\) \(2x=2\) \(x=2:2\) \(x=1\) | \(\left(2x-1\right)^2=1\) \(\left(2x-1\right)^2=\left(-1\right)^2\) \(\left(2x-1\right)=\left(-1\right)\) \(\Rightarrow2x-1=\left(-1\right)\) \(2x=\left(-1\right)+1\) \(2x=0\) \(x=0:2\) \(x=0\) |
\(x=1;0\)
HT
( 2x - 1 )2 + 17 = 18
( 2x - 1 )2 = 18 - 17
( 2x - 1 )2 = 1
( 2x - 1 )2 + 17 = 18
( 2x - 1 )2 = 18 - 17
( 2x - 1 )2 = 1
( 2x - 1 ) = 12
( 2x - 1 ) = 1
2x = 1 + 1
2x = 2
x = 2:2
x = 1
