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a)\(A=\left\{\left(0,1+0,9\right)+\left(0,2+0,8\right)+\left(0,3+0,7\right)+\left(0,4+0,6\right)+0,5\right\}\)
\(A=4.\left(1\right)+0,5=4,5\)
\(B=\left\{\left(1,01+1,09\right)+...\left(1,04+1,06\right)+1,05\right\}+1,1\)
\(B=4.\left(2,1\right)+1,05+1,1=8,4++1,05+1,1=10,46\)
A+B=4,5+10,46=14,96
\(=\frac{120-0,5\cdot40\cdot5\cdot0,2\cdot20\cdot0,25-20}{1+5+9+.....+33+37}\)
\(=\frac{120-20-\left(0,5\cdot20\right)\cdot\left(5\cdot0,2\right)\cdot\left(40\cdot0,25\right)}{1+5+9+.....+33+37}\)
\(=\frac{100-10\cdot1\cdot10}{1+5+9+.....+33+37}\)
\(=\frac{100-100}{1+5+9+.....+33+37}\)
\(=\frac{0}{1+5+9+.....+33+37}\)
\(=0\)
Ta có:\(120-0,5\times40\times5\times0,2\times20\times0,25-20=120-100-20=0\)
\(\Rightarrow\frac{120-0,5\times40\times5\times0,2\times20\times0,25}{1+5+.......+37}=\frac{0}{1+5+9+.....+37}=0\)
Vậy...........
\(27\cdot25\cdot\left(3+1\right)-230=27\cdot25\cdot4-230\)
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...+128+256}\)
\(=\frac{100-\left(20\times5\right)\times\left(0,5\times0,2\times0,25\times40\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-100\times\left(\frac{1}{40}\times40\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100\times\left(1-1\right)}{1+2+4+8+...+128+256}\)
\(=0\)
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...128+256}\)
\(=\frac{100-\left(0,5\times20\right)\times\left(40\times0,25\right)\times\left(5\times0,2\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-10\times10\times1}{1+2+4+8+...+128+256}\)
\(=\frac{100-100}{1+2+4+8+...+128+256}\)
\(=\frac{0}{1+2+4+8+...+128+256}=0\)
120 - 0,5 x 40 x 5 x 0,2 x 20 x 0,25 - 20
= ( 120 - 20 ) - [ ( 0,5 x 20 ) x ( 40 x 0,25 ) x ( 0,2 x 5 ) ]
= 100 - [ 10 x 10 x 1 ]
= 100 - 100
= 0
\(120-0,5\times40\times5\times0,2\times20\times0,25-20\)
\(=120-\left(0,5\times20\right)\times\left(40\times0,25\right)\times\left(5\times20\right)-20\)
\(=120-10\times10\times100-20\)
\(=120-10000-20\)
\(=-9900\)