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1/
a/ \(x^2+\left(y-10\right)^2=0\)
vì: \(\left\{{}\begin{matrix}x^2\ge0\forall x\\\left(y-10\right)^4\ge0\forall y\end{matrix}\right.\)
=> Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y-10=0\Rightarrow y=10\end{matrix}\right.\)
vậy......
b/ \(\left(0,5x-5\right)^{20}+\left(y^2-0,25\right)^{10}\le0\)
vì: \(\left\{{}\begin{matrix}\left(0,5x-5\right)^{20}\ge0\forall x\\\left(y^2-0,25\right)^2\ge0\forall y\end{matrix}\right.\)=> \(\left(0,5x-5\right)^{20}+\left(y^2-0,25\right)^{10}\ge0\)
=> Dấu ''='' xảy ra khi :
\(\left\{{}\begin{matrix}0,5x-5=0\\y^2-0,25=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{0,5}=10\\y^2=0,25\Rightarrow\left[{}\begin{matrix}y=0,5\\y=-0,5\end{matrix}\right.\end{matrix}\right.\)
Vậy........
2/ Ta có: \(2011\equiv1\left(mod10\right)\)
\(2011^{201}\equiv1^{201}\equiv1\left(mod10\right)\);
Có: \(1997^3\equiv3\left(mod10\right)\)
\(\left(1997^3\right)^4\equiv3^4\equiv1\left(mod10\right)\)
\(\left(1997^{12}\right)^{14}\equiv1^{14}\equiv1\left(mod10\right)\) hay \(1997^{168}\equiv1\left(mod10\right)\)
=> \(2011^{201}-1997^{168}\equiv1-1\equiv0\left(mod10\right)\)
hay \(2011^{201}-1997^{168}\) chia hết cho 10
=> Đpcm
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(\left|x-y-1\right|+\left|x+2\right|\ge0\)
Do \(\left|x-y-1\right|,\left|x+2\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-1=0\\x+2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
a: Ta có: \(x^2\ge0\forall x\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)
Do đó: \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(0;\dfrac{1}{10}\right)\)
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
a)(2x-3)2=1<=> \(\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}< =>\orbr{\begin{cases}2x=4\\2x=2\end{cases}}}\)\(< =>\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
x=2 =>22.52=20y.5y <=>100 = 100y <=> y=1
x=1 => 2.5= 20y.5y <=>10=100y <=>y = 1/2
b)(4x-3)2+(y2-9)2\(\ge0\)
dấu = sảy ra khi \(\hept{\begin{cases}4x-3=0\\y^2-9=0\end{cases}< =>\hept{\begin{cases}4x=3\\y^2=9\end{cases}}}\)\(\hept{\begin{cases}x=\frac{3}{4}\\y=\pm3\end{cases}}\)
c) <=> (y-5)8 \(\le-\left(x+4\right)^7\) (1)
(y-5)8 >=0 với mọi y nên -(x+4)7 \(\ge\left(y-5\right)^8\ge0\)<=> (x+4)7\(\le0< =>x+4\le0< =>x\le-4\)
Khi đó (1) <=> y-5\(\le\sqrt[8]{-\left(x+4\right)^7}\) <=> y\(\hept{\begin{cases}y\le5-\sqrt[8]{-\left(x+4\right)^7}\\x\le-4\end{cases}}\)
