Bài 1 :
a ) Ta có :
\(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)
\(=15^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
b ) Ta có :
\(x=11\Rightarrow x+1=12\)
Thay \(x+1=12\) vào biểu thức ta được :
\(x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+111\)
\(=x^4-x^4-x^3+x^3-x^2+x^2-x+111\)
\(=111-x\)
Thay \(x=11\) vào biểu thức vừa rút gọn ta được :
\(111-11=100\)
\(a,3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)
\(=\left(3.5\right)^4-\left(15^4-1\right)\)
\(=15^4-15^4+1\)
\(=1\)
Bài 2:
\(a,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2.\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(b,3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
