a)Đặt \(A=1^2+3^2+5^2+...+2023^2\)
Biểu thức \(P=A\cdot\left(4^3-8^2\right)=0\cdot A=0\)
b)\(K=\dfrac{2,5\cdot\left(0,8+1,2\right)}{15}+\dfrac{4\cdot\left(33,81-34,06\right)}{\left(-28,75+25,15\right):0,6}-0,375\)
\(K=\dfrac{2,5\cdot2}{15}+\dfrac{4\cdot\left(-0,25\right)}{\left(-3,6\right):0,6}-0,375\)
\(K=\dfrac{5}{15}+\dfrac{1}{6}-\dfrac{3}{8}=\dfrac{40+20-45}{120}=\dfrac{15}{120}=0,125\approx0,1\)
c)\(A=\dfrac{2022}{50^{10}}+\dfrac{2022}{50^8}=\dfrac{2022}{50^{10}}+\dfrac{2021}{50^8}+\dfrac{1}{50^8}\)
\(B=\dfrac{2023}{50^{10}}+\dfrac{2021}{50^8}=\dfrac{2022}{50^{10}}+\dfrac{1}{50^{10}}+\dfrac{2021}{50^8}\)
Ta có: \(50^8< 50^{10}\Rightarrow\dfrac{1}{50^8}>\dfrac{1}{50^{10}}\)
Vậy \(A>B\)
