Áp dụng đlý Pytago vào tam giác ABC:

AC²=BC²+AB²

5²=4²+3²

5²=25

Vậy tam giác ABC là tam giác vuông tại B (dpcm)

<=>giá trị tuyệt đối (x+2013).3+3=2

<=>giá trị tuyệt đối (x+2013).3=1

<=>giá trị tuyệt đối (x+2013=1/3

<=>x=-6038/3

vậyx=-6038/3

mk nha bạn mk trả lời đầu tiên

nếu x >= 0 thì /x+2013/+/x+2014/+/x+2015/ > 2013+2014+2015 > 2. x ko thể >= 0

nếu x>=-2013 thì: -2013+2013-2013+2014-2013+2015 >2

nếu x=-2014 thì: 2014-2013+2014-2014+2015-x=2 => x=-2014

nếu x>=-2015 thì: 2015-2013+2015-2014+2015-2015=3 

=> x chỉ bằng -2014

\(\frac{1}{x\left(x+1\right)}=\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

=>\(\frac{1}{x}-\frac{1}{x+1}-\frac{1}{x}=\frac{1}{2011}\)

=>\(\frac{1}{x}-\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2011}\)

=>\(0-\frac{1}{x+1}=\frac{1}{2011}\)

=>\(-\frac{1}{x+1}=\frac{1}{2011}\)

=>-x+1=2011

=>-x=2011-1

=>-x=2010

=>x=-2010

Vậy x=-2010

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

<=>\(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

<=>\(-\frac{1}{x+1}=\frac{1}{2011}\)

<=>-x-1=2011

<=>x=-2012

Đáp số: \(x=-2012\)

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