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24 - 16(x - 1/2) = 23
=> 16(x - 1/2) = 24 - 23
=> 16(x - 1/2) = 1
=> x - 1/2 = 1/16
=> x = 1/16 + 1/2
=> x = 9/16
\(24-16(x-\frac{1}{2})=23\)
\(16(x-\frac{1}{2})=24-23\)
\(16(x-\frac{1}{2})=1\)
\(x-\frac{1}{2}=\frac{1}{16}\)
\(x=\frac{1}{16}+\frac{1}{2}\)
\(x=\frac{9}{16}\)
Vậy số thực x cần tìm là \(\frac{9}{16}\)
Chúc bạn hok tốt ~
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Đặt \(\left|x-12\right|=t\ge0\Rightarrow\left(\left|x-12\right|\right)^2=\left(x-12\right)^2=t^2\) thay vào đẳng thức ta được :
\(2014t+t^2=2013t\)\(\Leftrightarrow t^2+2014t-2013t=0\Leftrightarrow t^2+t=0\)
\(\Leftrightarrow t\left(t+1\right)=0\Rightarrow\orbr{\begin{cases}t=0\\t=-1\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-12\right|=0\Rightarrow x=12\)
Vậy \(x=12\)
\(2^{x+1}.3^y=12^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.4^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.2^{2x}\)
\(\Rightarrow\orbr{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+1=2x\\y=x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\\text{Vì y = x}\Rightarrow y=1\end{cases}}\)
\(x^2+\frac{2}{9}=\frac{2}{3}\)
\(x^2=\frac{4}{9}\)
\(x^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow x=\frac{2}{3}\)
Học tốt~
\((x-\dfrac{1}{2})^2-\dfrac{1}{2}=\dfrac{23}{12}\)
\((x-\dfrac{1}{2})^2=\dfrac{23}{12}+\dfrac{1}{2}\)
\((x-\dfrac{1}{2})^2=\dfrac{29}{12}\)
\(\left[{}\begin{matrix}x-\dfrac{1}{2}=\sqrt{\dfrac{29}{12}}\\x-\dfrac{1}{2}=-\sqrt{\dfrac{29}{12}}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\sqrt{\dfrac{29}{12}}+\dfrac{1}{2}\\x=-\sqrt{\dfrac{29}{12}}+\dfrac{1}{2}\end{matrix}\right.\)