Question

a) \(\frac{4}{1x2}\) + \(\frac{4}{2x3}\)+ \(\frac{4}{3x4}\)+ ... + \(\frac{4}{2011x2012}\)

b) \(x\) x (\(x\)+ 1) =132

c) (1 + 4 + 7 + .... + 100) : x = 17

cảm ơnnnn

Answer 1

\(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2011.2012}\)

\(=4\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\right)\)

\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)

\(=4\left(1-\frac{1}{2012}\right)\)

\(=4.\frac{2011}{2012}\)

\(=\frac{2011}{503}\)

Answer 2

b. \(x.\left(x+1\right)=132\)

\(\Rightarrow x^2+x=132\)

\(\Leftrightarrow x=11\)

c. \(\left(1+4+7+...+100\right):x=17\)

\(\Rightarrow\frac{\left(100+1\right).34}{2}=17x\)

\(\Rightarrow1717=17x\)

\(\Rightarrow x=101\)