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34+25x=2960⇒25x=2960−34⇒25x=2960−3.1560⇒25x=29−4560⇒25x=−1660=−415⇒x=−415:25⇒x=−415.52⇒x=−23Vậyx=−23
\(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{40}=\dfrac{y}{24};\dfrac{y}{8}=\dfrac{z}{5}\Rightarrow\dfrac{y}{24}=\dfrac{z}{15}\\ \Rightarrow\dfrac{x}{40}=\dfrac{y}{24}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{40}=\dfrac{y}{24}=\dfrac{z}{15}=\dfrac{x+y+z}{40+24+15}=\dfrac{15,8}{79}=\dfrac{1}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=8\\y=\dfrac{24}{5}=4,8\\z=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{3}\\\dfrac{y}{8}=\dfrac{z}{5}\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{24}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{40}=\dfrac{y}{24}=\dfrac{z}{15}=\dfrac{x+y+z}{40+24+15}=\dfrac{15,8}{79}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}.40=8\\y=\dfrac{1}{5}.24=\dfrac{24}{5}\\z=\dfrac{1}{5}.15=3\end{matrix}\right.\)
\(N=\left|x-1004\right|+\left|x+1003\right|=\left|1004-x\right|+\left|x+1003\right|\le\left|1004-x+x+1003\right|=2007\)
Dấu "=" xảy ra khi \(\left(1004-x\right)\left(x+1003\right)\ge0\Leftrightarrow-1003\le x\le1004\)
Vậy MaxN = 2007 khi \(-1003\le x\le2004\)
N = |1004-x|+|x+1003| >= |1004-x+x+1003| = 2007
Dấu "=" xảy ra <=> (1004-x).(x+1003) >= 0
<=> -1003 <= x <= 1004
Vậy Min N = 2007 <=> -1003 <= x <= 1004