với x<1=>| x- 1| + | x-2| + |x-3| + x ==1-x+2-x+3-x+x=6-2x=0=>x=3(loại)

với 1<x<2=>| x- 1| + | x-2| + |x-3| + x ==x-1+2-x+3-x+x=4=0(loại)

với 2<x<3=>| x- 1| + | x-2| + |x-3| + x ==x-1+x-2+3-x+x=3x=0=>x=0(loại)

với 3<x=>| x- 1| + | x-2| + |x-3| + x =x-1+x-2+x-3+x=4x-6=0=>x=3/2(loại)

vậy ko tìm được x

tôi ko bt

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2012^2}-1\right)\)

\(=-\frac{1.3.2.4...2011.2013}{2^2.3^2...2012^2}\)

\(=-\frac{\left(1.2...2011\right)\left(3.4...2013\right)}{\left(2.3...2012\right)\left(2.3...2012\right)}\)

\(=-\frac{2013}{2012.2}=\frac{-2013}{4024}\)

Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)

\(\Rightarrow\left|x-2\right|+\left|x-5\right|\ge\left|\left(x-2\right)+\left(5-x\right)\right|\)

\(=3\)

(Dấu "="\(\Leftrightarrow\left(x-2\right)\left(5-x\right)\ge0\)

\(\Leftrightarrow2\le x\le5\))

/x-2/+/x-5/=3
hay /x-2/+/5-x/=3
mà /x-2/+/5-x/>=/x-2+5-x/=/3/=3
dấu bằng khi :(x-2)(5-x)>=0
gải bpt ta dc 2<=x<=5

hok tốt nha, nhớ k mik nhé

tôi ko bt

a.

\(\frac{2x}{7}=\frac{3y}{2}\Rightarrow4x=21y\)

\(x-y=17\Rightarrow x=17+y\)

\(\Rightarrow4\left(17+y\right)=21y\Rightarrow68+4y=21y\Rightarrow17y=68\Rightarrow y=4\)

\(\Rightarrow x=17+y=17+4=21\)

b.

\(\frac{x}{2}=\frac{y}{5}\Rightarrow5x=2y\)

\(x\cdot y=40\Rightarrow x=\frac{40}{y}\)

\(\Rightarrow5\cdot\frac{40}{y}=2y\Rightarrow\frac{200}{y}=2y\Rightarrow2y^2=200\Rightarrow y=\pm10\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)