Tìm \[x\]:

b) \[\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{x\left( {x + 1} \right)}} = \frac{{2008}}{{2009}}\];

a) \(A = 1,25 + \left| {2,5 - x} \right|\)

Ta có: \(\left| {2,5 - x} \right| \ge 0\) với mọi \(x\)

Do đó, \(1,25 + \left| {2,5 - x} \right| \ge 0 + 1,25\) hay \(1,25 + \left| {2,5 - x} \right| \ge 1,25\).

Dấu “=” xảy ra khi \(\left| {2,5 - x} \right| = 0\) suy ra \(2,5 - x = 0\) khi \(x = 2,5\).

Vậy giá trị nhỏ nhất của \(A = 1,25\) khi \(x = 2,5\).

c) \[\frac{{x - 214}}{{86}} + \frac{{x - 132}}{{84}} + \frac{{x - 54}}{{82}} = 6\]

\[x\left( {\frac{1}{{86}} + \frac{1}{{84}} + \frac{1}{{82}}} \right) = 6 + \frac{{214}}{{86}} + \frac{{132}}{{84}} + \frac{{54}}{{82}}\]

\[x\left( {\frac{1}{{86}} + \frac{1}{{84}} + \frac{1}{{82}}} \right) = \left( {1 + \frac{{214}}{{86}}} \right) + \left( {2 + \frac{{132}}{{84}}} \right) + \left( {3 + \frac{{54}}{{82}}} \right)\]

\[x\left( {\frac{1}{{86}} + \frac{1}{{84}} + \frac{1}{{82}}} \right) = \frac{{300}}{{86}} + \frac{{300}}{{84}} + \frac{{300}}{{82}}\]

\[x\left( {\frac{1}{{86}} + \frac{1}{{84}} + \frac{1}{{82}}} \right) = 300\left( {\frac{1}{{86}} + \frac{1}{{84}} + \frac{1}{{82}}} \right)\]

\[x = 300\]

Vậy \[x = 300\].