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a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}\cdot\left(x-3\right)^{10}=0\)
\(\left(x-3\right)^{x+5}\cdot\left[1-\left(x-3\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{x+5}=0\\1-\left(x-3\right)^{10}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{4;2\right\}\end{cases}}\)
Vậy........
a)\(2019-\left|x-2019\right|=x\)
\(\Rightarrow2019-x=\left|x-2019\right|\)
=>\(\left|x-2019\right|=-\left(x-2019\right)\)
=>\(x-2019\le0\)
=>\(x\le2019\)
b) Vì \(\left(2x-1\right)^{2018}\ge0\forall x\)
\(\left(y-\frac{2}{5}\right)^{2018}\ge0\forall y\)
\(\left|x+y-z\right|\ge0\forall x,y,z\)
=> \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|\ge0\forall x,y,z\)
mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}\)=>\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)
a, Ta có:
\(\left|x-2019\right|=\orbr{\begin{cases}x-2019\ge0\Rightarrow x\ge2019\\-x+2019< 0\Rightarrow x< 2019\end{cases}}\)
Xét x<2019 thì |x-2019|=-x+2019
Khi đó: 2019-(-x+2019)=x
\(\Leftrightarrow\)-x+2019=2019-x
\(\Leftrightarrow\)-x+2019+x=2019
\(\Leftrightarrow\)0x+2019=2019
\(\Leftrightarrow\)0x=0 (thỏa mãn)
Xét 2019\(\le\)x thì |x-2019|=x-2019
Khi đó 2019-(x-2019)=x
\(\Leftrightarrow\)2019-x+2019=x
\(\Leftrightarrow\)4038-x=x
\(\Leftrightarrow\)4038=2x
\(\Leftrightarrow\)x=2019(thỏa mãn)
Vậy .......................................................!!!
Vì : (3x+1)2018+(2y-1)2018+\(\left|x+2y-z\right|\)2018=0
Nên: \(\left\{{}\begin{matrix}\left(3x+1\right)^{2018}=0\\\left(2y-1\right)^{2018}\\\left|x+2y-z\right|^{2018}=0\end{matrix}\right.=0\) ⇔\(\left\{{}\begin{matrix}3x+1=0\\2y-1=0\\x+2y-z=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\\dfrac{-1}{3}+1-z=0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{matrix}\right.\)
Vậy : x=\(\dfrac{-1}{3}\) ; y=\(\dfrac{1}{2}\) ; z=\(\dfrac{2}{3}\)
\(\text{a) }\left(x-1\right)^2+\left|y+3\right|=0\)
Vì \(\left(x-1\right)^2\text{ và }\left|y+3\right|\text{ đều }\ge0\)
nên để \( \left(x-1\right)^2+\left|y+3\right|=0\)
thì \(\left(x-1\right)^2=0\text{ và }\left|y+3\right|=0\)
\(\Rightarrow x-1=0\text{ và }y+3=0\)
\(\Rightarrow x=1\text{ và }y=-3\)
\(\text{b) }\left(x^2-9\right)^2+\left|2-6y\right|^5\le0\)
\(\text{vì }\left(x^2-9\right)^2\text{ và }\left|2-6y\right|^5\text{ đều }\ge0\)
Nên để \(\left(x^2-9\right)^2+\left|2-6y\right|^5\le0\)
Thì \(\left(x^2-9\right)^2+\left|2-6y\right|^5=0\)
hay \(\left(x^2-9\right)^2=0\text{ và }\left|2-6y\right|^5=0\)
\(\Rightarrow x^2-9=0\text{ và }2-6y=0\)
\(\Rightarrow x^2=9\text{ và }6y=2\)
\(\Rightarrow x=\pm3\text{ và }y=\frac{1}{3}\)
Câu c) làm tương tự nha
b) \(\left|x-2018y\right|+\left(y-1\right)^{2018}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2018y\right|=0\\\left(y-1\right)^{2018}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018y=0\\y-1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018y=0\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018.1=0\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018=0\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2018\\y=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=2018\\y=1\end{matrix}\right.\)
c) \(\left|x+5\right|+\left(3y-4\right)^{2018}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2018}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\3y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
giúp mk lun con d) nha:
d) (2x-1)^2 +\(|2y-x|-8=12-5.2^2\)