a, \(\left(2x-1\right)=-8\)
\(2x=-8+1\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
a) (2x - 1) = -8
⇒ 2x = -8 + 1
⇒ 2x = -7
b) (3x - 2)\(^2\) = \(\frac{1}{49}\)
Ta có: \(\frac{1}{49}\) = \(\frac{1}{7}\). \(\frac{1}{7}\) hoặc \(\frac{1}{49}\) = \(\frac{-1}{7}\). \(\frac{-1}{7}\)
TH1: 3x - 2 = \(\frac{1}{7}\) TH2: 3x - 2 = \(\frac{-1}{7}\)
⇒ 3x = \(\frac{1}{7}\)+2 ⇒ 3x = \(\frac{-1}{7}\)+2
⇒ 3x = \(\frac{15}{7}\) ⇒ 3x = \(\frac{13}{7}\)
⇒ x = \(\frac{5}{7}\) ⇒ x = \(\frac{13}{21}\)
Vậy: x = \(\frac{5}{7}\) hoặc x = \(\frac{13}{21}\)
d, \(\frac{x^2}{6}=\frac{24}{25}\)
\(\Rightarrow x^2.25=24.6\)
\(\Rightarrow x^2.25=144\)
\(\Rightarrow x^2=144:25\)
\(\Rightarrow x^2=5,76\)
\(\Rightarrow x=2,4\) hoặc \(-2,4\)
b, \(\left(3x-2\right)^2=\frac{1}{49}\)
\(\left(3x-2\right)^2=\left(\frac{1}{7}\right)^2\)
\(3x-2=\frac{1}{7}\)
\(3x=\frac{1}{7}+2\)
\(3x=\frac{15}{7}\)
\(x=\frac{15}{7}:3\)
\(x=\frac{5}{7}\)
Có bn nào trả lời hộ mik nốt câu e với
a) \(\left(2x-1\right)=-8\)
⇒ \(2x=\left(-8\right)+1\)
⇒ \(2x=-7\)
⇒ \(x=\left(-7\right):2\)
⇒ \(x=-\frac{7}{2}\)
Vậy \(x=-\frac{7}{2}.\)
b) \(\left(3x-2\right)^2=\frac{1}{49}\)
⇒ \(3x-2=\pm\frac{1}{7}\)
⇒ \(\left[{}\begin{matrix}3x-2=\frac{1}{7}\\3x-2=-\frac{1}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3x=\frac{1}{7}+2=\frac{15}{7}\\3x=\left(-\frac{1}{7}\right)+2=\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{15}{7}:3\\x=\frac{13}{7}:3\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=\frac{5}{7}\\x=\frac{13}{21}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{7};\frac{13}{21}\right\}.\)
d) \(\frac{x^2}{6}=\frac{24}{25}\)
⇒ \(x^2.25=24.6\)
⇒ \(x^2.25=144\)
⇒ \(x^2=144:25\)
⇒ \(x^2=\frac{144}{25}\)
⇒ \(\left[{}\begin{matrix}x=\frac{12}{5}\\x=-\frac{12}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{12}{5};-\frac{12}{5}\right\}.\)
Chúc bạn học tốt!
