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\(\left(|x|-2017\right)^{\left(n+2018\right)\cdot\left(n+2019\right)}=-\left(2^3-3^2\right)^{2019}\)
\(\left(\left|x\right|-2017\right)^{\left(n+2018\right)\left(n+2019\right)}=-\left(2^3-3^2\right)^{2019}\)
\(\left(\left|x\right|-2017\right)^{\left(n+2018\right)\left(n+2019\right)}=-\left(-1\right)^{2019}=1\)
\(\Rightarrow\orbr{\begin{cases}\left(n+2018\right)\left(n+2019\right)=0\\\left|x\right|-2017=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\orbr{\begin{cases}n=-2018\\n=-2019\end{cases}}\\\orbr{\begin{cases}x=2018\\x=-2018\end{cases}}\end{cases}}\)
Từ \(\left(x+1\right)^6+\left(y-1\right)^4=-z^2\)
\(\Rightarrow\left(x+1\right)^6+\left(y-1\right)^4+z^2=0\)
Thấy: \(\left\{{}\begin{matrix}\left(x+1\right)^6\ge0\forall x\\\left(y-1\right)^4\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^6+\left(y-1\right)^4+z^2\ge0\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left(x+1\right)^6=0\\\left(y-1\right)^4=0\\z^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\\z=0\end{matrix}\right.\)
Khi đó \(N=2018\cdot x^{2016}\cdot y^{2017}-\left(z-1\right)^{2018}\)
\(=2018\cdot\left(-1\right)^{2016}\cdot1^{2017}-\left(0-1\right)^{2018}\)
\(=2018-\left(-1\right)^{2018}=2018-1=2017\)
thanks bạn nhiều nha Ace Legona. Mk cũng đang cần bài này
Ta có:
\(VT=\left|x-2017\right|+\left|2019-x\right|+\left|2018-x\right|\)
\(\Rightarrow VT\ge\left|x-2017+2019-x\right|+\left|2018-x\right|\)
\(\Rightarrow VT\ge2+\left|2018-x\right|\ge2\)
Dấu "=" xảy ra khi và chỉ khi \(x=2018\Rightarrow\) pt có nghiệm duy nhất \(x=2018\)
\(A=\left|x-2017\right|+\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)
\(\Rightarrow A=\left|x-2017\right|+\left|x-2018\right|+\left|2019-x\right|+\left|2020-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2017\right|+\left|x-2018\right|+\left|2019-x\right|+\left|2020-x\right|\ge\left|x-2017+x-2018+2019-x+2020-x\right|\)
\(\Rightarrow A\ge\left|4\right|\)
\(\Rightarrow A\ge4.\)
Dấu '' = '' xảy ra khi:
\(\left(x-2017\right).\left(x-2018\right).\left(2019-x\right).\left(2020-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2017\ge0\\x-2018\ge0\\2019-x\ge0\\2020-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2017\le0\\x-2018\le0\\2019-x\le0\\2020-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2017\\x\ge2018\\x\le2019\\x\le2020\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2017\\x\le2018\\x\ge2019\\x\ge2020\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2018\le x\le2019\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_A=4\) khi \(2018\le x\le2019.\)
Chúc bạn học tốt!
