Tính a^4+b^4 biết a+b=10;ab=4
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#include <iostream>
using namespace std;
int main() {
long long a, b, c;
cin >> a >> b >> c;
long long result = ((a % c) + (b % c)) % c;
cout << result << endl;
return 0;
}
a, b = map(int, input().split())
c = int(input())
result = ((a % c) * (b % c)) % c
print(result)
\(S=1.3^0+2.3^1+3.3^2+...+11.3^{10}\)
\(3S=1.3^1+2.3^2+...+11.3^{11}\)
\(\Rightarrow S-3S=1+3^1+3^2+...+3^{10}-11.3^{11}\)
\(\Rightarrow-2S=1.\dfrac{3^{11}-1}{3-1}-11.3^{11}\)
\(\Rightarrow-2S=\dfrac{1}{2}.3^{11}-\dfrac{1}{2}-11.3^{11}\)
\(\Rightarrow-2S=-\dfrac{21.3^{11}+1}{2}\)
\(\Rightarrow S=\dfrac{1}{4}+\dfrac{21.3^{11}}{4}\)
Bài 1:
a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)
\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{6}{7}\)
b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)
\(\Rightarrow x=\dfrac{1}{6}\)
a) Đặt A=1/2 + 1/4 + 1/8 +...+ 1/256 + 1/512
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
b)\(\frac{a}{b}+\frac{4}{6}+\frac{2}{10}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b}+\frac{13}{15}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b}=\frac{19}{30}\)
\(\frac{4}{5}:\frac{a}{b}-\frac{6}{5}=\frac{3}{10}\)
\(\Rightarrow\frac{4}{5}:\frac{a}{b}=\frac{3}{2}\)
\(\Rightarrow\frac{a}{b}=\frac{8}{15}\)
(a+b)2 = 100, 2ab = 8 ⇔ a2 + b2 = 100 - 8 = 92
(a2 + b2)2 = 922 = 8464 , ab = 4 ⇒ a2b2 = 16 ⇒ 2a2b2 = 32
a4 + b4 = 8464 - 32 = 8432
\(=\left(a^2+b^2\right)^2-2a^2b^2=\left[\left(a+b\right)^2-2ab\right]-2a^2b^2\)
-> \(\left(100-8\right)-2.16=92-32=60\)