đặt biểu thức \(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\) là A
Ta có:\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=\left(4^{16}-1\right)\left(4^{16}+1\right)\)
\(\Rightarrow15.A=4^{32}-1\)
\(\Rightarrow A=\dfrac{4^{32}-1}{15}\)
Vậy giá trị biểu thức trên là \(\dfrac{4^{32}-1}{15}\)
\(b,40^2-39^2+38^2-37^2+...+2^2-1^2\)
\(=\left(40^2-39^2\right)+\left(38^2-37^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(40-39\right)\left(40+39\right)+\left(38-37\right)\left(38+37\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=1+2+...+38+39+40\)
\(=\dfrac{\left(40+1\right).40}{2}=\dfrac{41.40}{2}=820\)
