Gọi số lít xăng E5 và số lít xăng A95 lần lươt là a,b (lít) \(\left(0< a;b< 1000\right)\)

Theo bài ra, ta có: \(\hept{\begin{cases}a+b=1000\\15000a+17000b=16300000\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}15a+15b=15000\\15a+17b=16300\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(15a+17b\right)-\left(15a+15b\right)=16300-15000\\a+b=1000\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=650\\a=350\end{cases}}\) (thỏa mãn)

Vậy trạm đó bán được 350 lít xăng E5 và 650 lít xăng A95

Ta có: \(\left(\frac{8}{1.9}+\frac{8}{9.17}+\frac{8}{17.25}+...+\frac{8}{49.57}\right)+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)

\(\Leftrightarrow1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+\frac{1}{17}-\frac{1}{25}+....+\frac{1}{49}-\frac{1}{57}+2x-2=\frac{8x+28+15x-24}{12}\)

\(\Leftrightarrow1-\frac{1}{57}+2x-2=\frac{23x+4}{12}\)

\(\Leftrightarrow2x-\frac{58}{57}=\frac{23x+4}{12}\)

\(\Leftrightarrow24x-\frac{232}{19}=23x+4\)

\(\Leftrightarrow x=\frac{308}{19}\)

\(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\)

\(=\frac{xyz}{x\left(1+y+yz\right)}+\frac{1}{1+y+yz}+\frac{xyz}{xz\left(1+y+yz\right)}\)

\(=\frac{yz}{1+y+yz}+\frac{1}{1+y+yz}+\frac{y}{1+y+yz}\)

\(=\frac{1+y+yz}{1+y+yz}\)

\(=1\)

\(\frac{x-15}{2014}+\frac{x-20}{2019}=\frac{x-5}{2004}+\frac{x+30}{1969}\)

\(\Leftrightarrow\frac{x-15}{2014}+1+\frac{x-20}{2019}+1=\frac{x-5}{2004}+1+\frac{x+30}{1969}+1\)

\(\Leftrightarrow\frac{x-15+2014}{2014}+\frac{x-20+2019}{2019}-\frac{x-5+2004}{2004}-\frac{x+30+1969}{1969}=0\)

\(\Leftrightarrow\frac{x-1999}{2014}+\frac{x+1999}{2019}-\frac{x+1999}{2004}-\frac{x+1999}{1969}=0\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{2014}+\frac{1}{2019}-\frac{1}{2004}-\frac{1}{1969}\right)=0\)

Vì \(\left(\frac{1}{2014}+\frac{1}{2019}-\frac{1}{2004}-\frac{1}{1969}\right)\ne0\)

nên \(x-1999=0\)

\(\Leftrightarrow x=1999\)

\(easy!\)(sai đề + sửa đề)

\(\frac{x-5}{2014}+\frac{x-20}{2019}-\frac{x-5}{2004}-\frac{x+3}{1969}=0\)

\(\Leftrightarrow\left(\frac{x-15}{2014}-1\right)+\left(\frac{x-20}{2019}-1\right)-\left(\frac{x-5}{2004}-1\right)-\left(\frac{x-30}{1969}-1\right)=0\)

\(\Leftrightarrow\frac{x-1999}{2014}+\frac{x-1999}{2019}-\frac{x-1999}{2004}-\frac{x-1999}{1969}=0\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{2014}+\frac{1}{2019}-\frac{1}{2004}-\frac{1}{1969}\right)=0\)

dễ dàng cm được \(x-1999=0\)

\(\Leftrightarrow x=1999\)

Sai đề kìa

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\ge\frac{2}{\frac{x+y}{2}}=\frac{4}{x+y}\)

Dấu " = " xảy ra \(\Leftrightarrow\frac{1}{x}=\frac{1}{y}\Leftrightarrow x=y\)

haiz!chán vcl nên mới trả lời câu này

Áp dụng bất đẳng thức AM-GM,ta có:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

dấu "=" xảy ra khi và chỉ khi \(x=y\)

\(\Rightarrow you\)sai đề