a) 

b) 

c) 

d) 

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)

\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)

\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)

\(x\times4+\frac{40}{81}=\frac{51}{81}\)

\(x\times4=\frac{51}{81}-\frac{40}{81}\)

\(x\times4=\frac{11}{81}\)

\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)

\(\Rightarrow x=\frac{11}{324}\)

[ 61 + ( 53 - x ) ] . 17 = 1785

61 + ( 53 - x ) = 1785 : 17

61 + ( 53 - x ) = 105

( 53 - x ) = 105 - 61

53 - x = 44

=> x = 53 - 44

=> x = 9

a) \(\left( { - 3} \right).7 =  - \left( {3.7} \right) =  - 21\)

b) \(\left( { - 8} \right).\left( { - 6} \right) = 8.6 = 48\)

c) \(\left( { + 12} \right).\left( { - 20} \right) =  - \left( {12.20} \right) =  - 240\)

d) \(24.\left( { + 50} \right) = 24.50 = 1200\)

a,\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)

\(=\frac{-8}{9}.\frac{-9}{10}...\frac{-2003}{2004}.\frac{-2004}{2005}\)

\(=\frac{\left(-8\right).\left(-9\right)...\left(-2003\right).\left(-2004\right)}{9.10...2004.2005}\)

\(=\frac{-\left(8.9...2003.2004\right)}{9.10...2004.2005}\)

\(=\frac{-8}{2005}\)

b,Ta có: \(81^{10}-27^{13}-9^{21}\)

\(=\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)

\(=3^{40}-3^{39}-3^{42}\)

\(=3^{39}.3-3^{39}-3^{39}.3^3\)

\(=3^{39}.\left(3-1-3^3\right)\)

\(=3^2.3^{37}.\left(-25\right)\)

\(=3^{37}.\left(-225\right)⋮225\)

Vậy \(81^{10}-27^{13}-9^{21}⋮225\)

a)\({\left( { - 2} \right)^2}.{\left( { - 2} \right)^3} = {\left( { - 2} \right)^{2 + 3}} = {\left( { - 2} \right)^5}\);

b)\({\left( { - 0,25} \right)^7}:{\left( { - 0,25} \right)^5} = {\left( { - 0,25} \right)^{7 - 5}} = {\left( { - 0,25} \right)^2} = {\left( {0,25} \right)^2}\);

c)\({\left( {\frac{3}{4}} \right)^4}.{\left( {\frac{3}{4}} \right)^3} = {\left( {\frac{3}{4}} \right)^{4 + 3}} = {\left( {\frac{3}{4}} \right)^7}.\)

a) \(\left( { - 3} \right).\left( { - 2} \right).\left( { - 5} \right).4\)\( = \left[ {\left( { - 3} \right).\left( { - 2} \right)} \right].\left( { - 5} \right).4\)\( = 6.\left( { - 5} \right).4 =  - 30.4 =  - 120\).

b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).

\(a,=\left(4x^2-25\right)-\left(6x^2+9x-4x-6\right)\\ =4x^2-25-6x^2-5x+6=-2x^2-5x-19\\ b,=4x^2-4x+1-4\left(x^2-4\right)\\ =4x^2-4x+1-4x^2+16\\ =-4x+17\)

\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)

a: \(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

b: \(=\dfrac{2x^4-2x^3+2x^2+3x^3-3x^2+3x-2x^2+2x-2-x-1}{x^2-x+1}\)

\(=2x^2+3x-2+\dfrac{-x-1}{x^2-x+1}\)