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\(A=\frac{5^{60}+1}{5^{61}+1}\)
\(5A=\frac{5(5^{60}+1)}{5^{61}+1}=\frac{5^{61}+5}{5^{61}+1}=\frac{5^{61}+1+4}{5^{61}+1}=1+\frac{4}{5^{61}+1}\) \((1)\)
\(B=\frac{5^{61}+1}{5^{62}+1}\)
\(5B=\frac{5(5^{61})+1}{5^{62}+1}=\frac{5^{62}+5}{5^{62}+1}=\frac{5^{62}+1+4}{5^{62}+1}=1+\frac{4}{5^{62}+1}\) \((2)\)
Từ 1 và 2 \(\Rightarrow1+\frac{4}{5^{61}+1}>1+\frac{4}{5^{62}+1}\)
\(\Rightarrow5A>5B\)
Hay \(A>B\)
Vậy : ...
a)\(568-\left\{5.\text{[}143-\left(4-1\right)^2\text{]}+10\right\}:10\)
\(=568-\left\{5.\left[143-3^2\right]+10\right\}:10\)
\(=568-\left\{5.134+10\right\}\text{ }\)
\(=568-\left(670+10\right)\)
\(=568-680\)
\(=-112\)
b)\(10^2-\left[60:\left(5^6:5^4-3\times5\right)\right]\)
\(=100-\left[60-\left(6^2-15\right)\right]\)
\(=100-\left(60-\left(36-15\right)\right)\)
\(=100-60-36+15\)
\(=19\)
\(a)\)\(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
\(b)\)\(10^2-\left[60\div\left(5^6\div5^4-3\times5\right)\right]\)
\(=\)\(10^2-\left[60\div\left(5^2-15\right)\right]\)
\(=\)\(10^2-\left[60\div\left(25-15\right)\right]\)
\(=\)\(10^2-\left[60\div10\right]\)
\(=\)\(100-6\)
\(=\)\(94\)
\(2b)\)
Đặt :
\(S=1+4+4^2+4^3+4^4....................+4^{100}\)
\(4S=4\left(1+4+4^2+4^3+4^4+.............+4^{100}\right)\)
\(4S=4+4^2+4^3+4^4+4^4+.......+4^{101}\)
\(4S-S=\left(4+4^2+4^3+4^4+4^5+.......+4^{101}\right)-\left(1+4+4^2+4^3+4^4+...............+4^{100}\right)\)
\(3S=4^{101}-1\)
\(S=\dfrac{4^{101}-1}{3}\)
A = 23 . 19 - 23 . 14 + 12018
= 23.(19 - 14) + 1
= 8 . 5 + 1
= 40 + 1 = 41
B = 102 - [60 : (56 : 54 - 3. 5)]
= 100 - [60 : (10 - 15)]
= 100 - [60 : (-5)]
= 100 + 12
= 112
nhân 5A
sau đs tính 5A-A=4A
=> A=......
ss vs B=5^61/4