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Ta có \(|x-y+3|\ge0\forall x,y\)
\(2015\left(2y-3\right)^{2016}\ge0\forall y\)
\(\Rightarrow\hept{\begin{cases}|x-y+3|\ge0\\2015.\left(2y-3\right)^{2016}\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\\left(2y-3\right)^{2016}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\2y-3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\2y=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+3=0\\y=\frac{3}{2}\end{cases}}\)
Bạn thay vào tìm x
Mik cũng hok Toán 2
\(a)2018=\left|x-2016\right|+\left|x-2014\right|\)
\(\Rightarrow\hept{\begin{cases}x-2016+x-2014=2018\\x-2016+x-2014=-2018\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-2016-2014=2018\\2x-2016-2014=-2018\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=2018+2016+2014\\2x=-2018+2016+2014\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=6048\\2x=2012\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3024\\x=1006\end{cases}}\)
vậy x = 3024 hoặc x = 1006
b) \(\left(x-3\right)^x-\left(x-3\right)^{x+2}=0\)
\(\Rightarrow\left(x-3\right)^x-\left(x-3\right)^x\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)^x\left[1-\left(x-3\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^x=0\\1-\left(x-3\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-3=0\\\left(x-3\right)^2=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\\left(x-3\right)^2=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\x-3=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=4\end{cases}}\)
vậy x = 3 hoặc x = 4
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(=>11-3x+1=\frac{9}{2}-5+3,5x\)
\(=>-3x+12=3,5x-\frac{1}{2}\)
\(=>-3x-3,5x=-\frac{1}{2}-12\)
\(=>-6,5x=-12,5\)
\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)
Ủng hộ nha
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(11-3x+1=\frac{9}{2}-5+3,5x\)
\(12-3x=-\left(0,5\right)+3,5x\)
\(12,5-3x=3,5x\)
\(12,5=6,5x\)
\(x=12,5:6,5=\frac{25}{13}\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)
\(\Leftrightarrow x=216-1=215\)
Ta có: \(3^x+3^{x+1}+3^{x+2}=351\)
\(\Rightarrow3^x.1+3^x.3+3^x.3^2^{ }=351\)
\(\Rightarrow3^x.1+3^x.3+3^x.9=351\)
\(\Rightarrow3^x.\left(1+3+9\right)=351\)
\(\Rightarrow3^x.13=351\)
\(\Rightarrow3^x=351:13=27\)
\(\Rightarrow x=3\)
3 k mk nha
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