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a) 17.23+17.77
=850
b) 12000 - ( 1500 .2 + 1800 .3 +1800 .2 : 3
=2400
c) [173. ( 17 +6 ) + 137.6^7 ]: 137
=279959
d) [2^10. ( 137 - 111 ) +2^10 .84)] : 1024: 5^2
=4,4
\(A=98.42-\left\{50.\left[\left(18-2^3\right):2+3^2\right]\right\}\)
\(=98.42-\left\{50.\left[\left(18-8\right):2+9\right]\right\}\)
\(=98.42-\left[50\left(10:2+9\right)\right]\)
\(=98.42-\left(50.14\right)\)
\(=4116-700=3416\)
\(B=-80-\left[-130-\left(12-4\right)^2\right]+2008^0\)
\(=-80-\left(-130-8^2\right)+1\)
\(=-80-\left(-130-64\right)+1\)
\(=-80+130+64+1\)
\(=115\)
\(C=1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(=1024:16+140:\left(38+32\right)-7^2\)
\(=64+140:70-49\)
\(=64+2-49=17\)
\(D=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).\left(16-16\right)\)
\(=\left(2^{17}+15^4\right).\left(3^{19}-2^{17}\right).0\)
\(=0\)
\(E=100+98+96+....+4+2-97-95-....-3-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+.....+\left(2-1\right)+\left(1-0\right)\)
\(=100+1+1+...+1+1\)
Vì lập được 49 cặp nên sẽ có 49 số 1
\(\Rightarrow E=100+1.49=100+49=149\)
a/A=1+2+4+8+...+1024
2A=2+4+8+16+....+2048
2A-A=(2+4+8+16+....+2048)-(1+2+4+8+...+1024)
A=2048-1
A=2047
VẬY A=2047
b/B=1+5+25+125+....+15625
5B=5+25+125+625+....+78125
5B-B=(5+25+125+625+....+78125)-(1+5+25+125+....+15625)
4B=78125-1
4B=78124
B=78124:4
B=19531
VẬY B =19531
C=1/1.2+1/2.3+1/3.4+...+1/2015.2016
C=1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016
=1-1/2016
=2015/2016
VẬY C=2015/2016
D/=10/1.3+10/3.5+10/5.7+....+10/2013.2015
=5(2/1.3+2/3.5+2/5.7+...+2/2013.2015)
=5(1-1/3+1/3-1/5+1/5-1/7+..+1/2013-1/2015)
=5(1-1/2015)
=5.2014/2015
=2014/403
VẬY D=2014/403
a, A = 1 + 2 + 4 + 8 +...+ 1024
\(A=1+2+2^2+2^3+....+2^{10}\)
\(2A=2+2^2+2^3+....+2^{10}+2^{11}\)
\(A=1+2+2^2+2^3+....+2^{10}\)
\(A=2^{11}-1=2047\)
b, B = 1 + 5 + 25 + 125 + ... + 15625
\(B=1+5+5^2+5^3+....+5^6\)
\(3B=5+5^2+5^3+....+5^6+5^7\)
\(B=1+5+5^2+5^3+....+5^6\)
\(2B=5^7-1\Rightarrow B=\frac{5^7-1}{2}=39062\)
d, D = 10 / 1 . 3 + 10 / 3 . 5 + 10 / 5 . 7 + ... + 10 / 2013 . 2015
\(D=\frac{10}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(D=\frac{10}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(D=\frac{10}{2}.\left(1-\frac{1}{2015}\right)=5.\frac{2014}{2015}=\frac{2014}{403}\)
Câu c thì tương tự
\(\frac{1}{2.5}\)\(+\)\(\frac{1}{5.8}\)\(+\frac{1}{8.11}\)\(+...+\frac{1}{152.155}\)
=\(\frac{1}{2}\) \(-\frac{1}{5}\) \(+\frac{1}{5}\) \(-\frac{1}{8}\) \(+...+\frac{1}{152}\) \(-\frac{1}{155}\)
=\(\frac{1}{2}\)\(-\frac{1}{155}\)
=\(\frac{153}{310}\)
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{1000}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right)...\left(-\frac{999}{1000}\right)\)
\(=-\frac{1.2.3...999}{2.3.4...1000}=-\frac{1}{1000}\)
b)\(B=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}:\frac{3}{4}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}:\frac{3}{4}=\frac{3}{4}:\frac{3}{4}=1\)
d) \(D=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
=> \(D=2-\frac{1}{1024}=\frac{2047}{1024}\)
1024:(26:25+42:4)
=1024 : ( 26-5+42-1 )
=1024 : ( 2+4 )
=1024:6 = 512/3
!!!!
x^20-x=0
x(x^19-1)=0
x= 0
hoặc x ^ 19 =1
x = 0 hoặc x= 1
\(a)\frac{1024}{17\cdot2^5+15\cdot2^5}=\frac{1024}{(17+15)\cdot2^5}=\frac{1024}{32\cdot32}=\frac{1024}{1024}=1\)
\(b)2^2\cdot102-2^3\cdot17-2^2\cdot34=2^2\cdot(102-34)-2^3\cdot17\)
\(=4\cdot68-8\cdot17\)
\(=272-136=136\)
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