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Bài 1 :
Vì \(\sqrt{3x+2y+z}\ge0\forall x;y;z\)
\(\left|y-\frac{1}{2}\right|\ge0\forall y\)
\(\left(z-2\right)^2\ge0\forall z\)
\(\Rightarrow A\ge2018\forall x;y;z\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x+2y+z=0\\y-\frac{1}{2}=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x+2\cdot\frac{1}{2}+2=0\\y=\frac{1}{2}\\z=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=\frac{1}{2}\\z=2\end{cases}}}\)
Vậy........
Bài 2 :
Lý luận tương tự câu 1) ta có :
\(\hept{\begin{cases}x-1=0\\y+1=0\\x+y+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\\1-1+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\\z=0\end{cases}}}\)
Thay x; y; z vào P ta có :
\(P=1^{2018}+\left(-1\right)^{2019}+0^{2020}\)
\(P=1-1+0\)
\(P=0\)
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}-1\right)+\left(\frac{x+2}{2018}-1\right)=\left(\frac{x+3}{2017}-1\right)+\left(\frac{x+4}{2016}-1\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2020=0:\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)\)
\(\Leftrightarrow x+2020=0\)
Còn lại tự làm :V
Lộn chỗ này , thay chút nha !
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)
Sorry =))
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+\frac{2}{2018}+\frac{3}{2017}+...+\frac{2018}{2}+\frac{2019}{1}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+1+\frac{2}{2018}+1+\frac{3}{2017}+1+...+\frac{2018}{2}+1+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2020}{2019}+\frac{2020}{2018}+\frac{2020}{2017}+...+\frac{2020}{2}+\frac{2020}{2020}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}\right)}\)
\(\frac{A}{B}=\frac{1}{2020}\)
\(x=\frac{2^{2018}+2^{2019}}{2^{2017}+2^{2018}}=\frac{2^{2019}}{2^{2017}}=\frac{2^2}{2}=\frac{4}{2}=2.\)
\(y=\sqrt{10^2-8^2}-\sqrt{3^2+4^2}\)
\(=\sqrt{100-64}-\sqrt{9+16}\)
\(=\sqrt{36}-\sqrt{25}\)
\(=6-5=1\)