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Đặt:
\(A=4.5^{100}.\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.....+\dfrac{1}{5^{100}}\right)+1\)
\(S=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.....+\dfrac{1}{5^{100}}\)
\(5S=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.....+\dfrac{1}{5^{100}}\right)\)
\(5S=1+\dfrac{1}{5}+\dfrac{1}{5^2}+.....+\dfrac{1}{5^{99}}\)
\(5S-S=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+.....+\dfrac{1}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{100}}\right)\)\(4S=1-5^{100}\Rightarrow S=\dfrac{1-5^{100}}{4}\)
Thay S và A ta có:
\(A=4.5^{100}.\dfrac{1-5^{100}}{4}+1\)
\(A=5^{100}.\left(1-5^{100}\right)+1\)
\(A=5^{100}-5^{200}+1\)
\(2b)\)
Đặt :
\(S=1+4+4^2+4^3+4^4....................+4^{100}\)
\(4S=4\left(1+4+4^2+4^3+4^4+.............+4^{100}\right)\)
\(4S=4+4^2+4^3+4^4+4^4+.......+4^{101}\)
\(4S-S=\left(4+4^2+4^3+4^4+4^5+.......+4^{101}\right)-\left(1+4+4^2+4^3+4^4+...............+4^{100}\right)\)
\(3S=4^{101}-1\)
\(S=\dfrac{4^{101}-1}{3}\)
(2x+24).53=4.55
=> 2x+24=4.55:53
=> 2x+16=4.52
=> 2x+16=4.25
=> 2x+16=100
=> 2x=100-16
=> x=84:2
=> x=42
(2x+24).53=4.55
=> 2x +2^4 = 4.5^5:5^3
=> 2x+16 = 4.5^2
=> 2x+16 = 4.25
=> 2x+16= 100
=> 2x= 100-16
x=84
=> x= 84:2
x=42
a) \(4.5^2-81:3^2=4.25-81:9=100-9=91\)
b) \(3^3.23-3^3.19=3^3.\left(23-19\right)=27.4=108\)
c) \(2^4.5-[131-\left(13-4\right)^2]=16.5-\left(131-9^2\right)=80-\left(131-81\right)=80-50=30\)
d) \(100:\left\{250:\left[450-\left(4.5^3-2^2.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-4.\left(125-25\right)\right]\right\}\)
\(=100:\left[250:\left(450-4.100\right)\right]\)
\(=100:\left[250:\left(450-400\right)\right]=100:\left(250:50\right)=100:5=20\)
100:{250:[450-(4.53-22.25)}]
=100:{250:[450-(500-100)]}
=100:{250:[450-400]}
=100:{250:50}
=100:5
=20
Là :
= 100 : {300 : [450 - (500-200)]}
= 100 : {300 : [450-300]}
= 100 : {300 : 150}
= 100 : 2
= 50
\(4+4\cdot5+4\cdot5^2+...+4\cdot5^{100}\\ =4\left(1+5+5^2+...+5^{100}\right)\left(1\right)\)
Đặt \(A=1+5+5^2+...+5^{100}\)
\(\Leftrightarrow5A=5+5^2+...+5^{101}\\ \Leftrightarrow4A=5^{101}-1\\ \Leftrightarrow A=\dfrac{5^{101}-1}{4}\)
Thay vào (1)
\(\left(1\right)=4\cdot\dfrac{5^{101}-1}{4}=5^{101}-1:5^{101}-1=1\)
Vậy \(4+4\cdot5+4\cdot5^2+...+4\cdot5^{100}:5^{101}-1=1\)