cau a dau nhi cuoi cung k phai j dau nha ! mk an lom ! 

\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)

 \(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)

\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)

ta có |x+5| \(\ge\)0 \(\forall x\)

Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn

b,

 \(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)

\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)

Ta có: \(1-\frac{1}{2}.\frac{2}{3}....\frac{98}{99}\)

 = \(1-\frac{1}{99}\)

=    \(\frac{98}{99}\)

P=\(1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).........\left(1-\frac{1}{99}\right)\)

P=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{98}{99}=\frac{1}{99}\)

\(P=1-\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Nha ban Speed Ninja

\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)

=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)

=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)

Đáp số: C=1

C=1

HT

C = 1

JY 

=\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2013}{2014}x\frac{2014}{2015}\)

=\(\frac{1x2x3x...x2013x2014}{2x3x4x...x2014x2015}\)

=\(\frac{1}{2015}\)

( Dau x la dau nhan)

\(\frac{2}{2015}\)

1/2015 dung day minh lam roi

a)

\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\=  - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)                          

b)

\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)                  

d)

\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ =  - \frac{3}{4}\end{array}\).

\(\left(1+\frac{1}{11}\right)\cdot\left(1+\frac{1}{10}\right)\cdot\left(1+\frac{1}{9}\right)\cdot.........................................\left(1+\frac{1}{2}\right)\)

\(=\frac{12}{11}.\frac{11}{10}.\frac{10}{9}....\frac{3}{2}\)

\(=\frac{12.11.10....3}{11.10.9....2}\)

\(=\frac{12}{2}=6\)

= \(\frac{12}{11}.\frac{11}{10}.....\frac{3}{2}=\frac{12}{2}=6\)

Dài thế!