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có 2 cách thông thường:
cách 1:
A=(100+2)x50:2-(97+1)x49:2
=2550-2401=149
cách 2:
A=100+(98-97)+(96-95)+...+(2-1)
=100+1+1+...+1( có 49 số 1)=100+49=149
\(=2^{100}-\left(2^{99}+2^{98}+2^{97}+...+2+1\right)\)
Đặt \(B=1+2+2^2+...+2^{98}+2^{99}\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow B=\left(2+2^2+2^3+..+2^{100}\right)-\left(1+2+2^2+...+2^{99}\right)\)
\(\Rightarrow B=2^{100}-1\)
\(\Rightarrow2^{100}-2^{99}-2^{98}-....-2-1=2^{100}-\left(2^{100}-1\right)\)
\(=1\)
\(\frac{99}{98}+\frac{96}{97}+\frac{1}{97.98}=\frac{99}{98}+\frac{96}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}+\frac{96}{97}\right)=1+1=2\)
= 2
mk bấm máy tính
đúng 10000000000000000000000000000000000000000000000000000000000000000000000%
\(A+1=2^{100}-2^{99}+2^{98}-...-2+1.\)
\(3\cdot\left(A+1\right)=\left(2+1\right)\left(2^{100}-2^{99}+2^{98}-...-2+1\right).\)
\(3\cdot\left(A+1\right)=2^{101}+1\)
\(A=\frac{1}{3}\cdot\left(2^{101}+1\right)-1=\frac{2^{101}-2}{3}\)
a) \(\frac{3^{10}.\left(11+5\right)}{3^9.16}\)=\(\frac{3^{10}.16}{3^{10}.16}\)=1
a) \(\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=\frac{3^{10}}{3^9}=3\)
\(96^{97^{98}}\times99^{100^{101}}\times103^{104^{105}}>96^{97^{98}}\times100^{101^{102}}\times102^{103^{104}}>97^{98^{99}}\times99^{100^{101}}\times102^{103^{104}}\)
a) \(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow3A=A+2A=2^{101}-2\)
\(\Rightarrow A=\frac{2^{101}-2}{3}\)
b) \(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
\(\Rightarrow4B=B+3B=3^{101}+1\)
\(\Rightarrow B=\frac{3^{101}+1}{4}\)
Bài 30 :
a ) Ta có :
( a + b ) ( a - b )
= ( a + b ) . a - ( a + b ) . b
= a . a + ab - ab - b . b
= a2 + ab - ab - b2
= a2 - b2 ( điều phải chứng minh )
b ) M = 1002 - 992 + 982 - 972 + 962 - 952 + ..... + 42 - 32 + 22 - 12
M = 199 + 195 + 191 + ...... + 7 + 3
M = ( 199 + 3 ) x [ ( 199 - 3 ) : 4 + 1 ] : 2
M = 202 x 50 : 2
M = 10100 : 2
M = 5050
30) Ta có : \(\left(a+b\right)\left(a-b\right)\)
\(=\left(a+b\right)a-\left(a+b\right).b\)
\(=a^2+ab-ab-b^2\)
\(=a^2-b^2\left(đpcm\right)\)
\(D=99-97+95-93+...+7-5+3-1\)
\(\Rightarrow D=\left(99-97\right)+\left(95-93\right)+...+\left(7-5\right)+\left(3-1\right)\) ( 25 cặp số )
\(\Rightarrow D=2+2+...+2+2\) ( 25 số 2 )
\(\Rightarrow D=2.25\)
\(\Rightarrow D=50\)
Vậy D = 50
A = 100 + 98 = 96 + …= 2 - 97 - 95 - …-1
Đặt M = 100 + 98 + 96 + … + 2; N = 97 + 95 + …+ 3+ 1
Nên M = ( 100 + 2) [ 100 - 2) :2 +1] : 2
=> M = 102.50 : 2 = 2550
Lại có: N = ( 97 +1). [ ( 97 - 1 ) : 2 +1 ]: 2
ð N = 98. 49 : 2 = 2401
Do đó : A = M = N = 149