Bài 1:
Thay x=6 vào biểu thức \(A=x^5-7x^4+7x^3-7x^2+1\), ta được:
\(A=6^5-7\cdot6^4+7\cdot6^3-7\cdot6^2+1\)
\(=7776-7\cdot1296+7\cdot216-7\cdot36+1\)
\(=7776-9072+1512-252+1\)
\(=-35\)
Vậy: Khi x=6 thì A=-35
Bài 2:
Ta có: \(\left(\frac{1}{2}x^2y+3\right)\cdot\left(\frac{1}{4}x^4y^2-\frac{3}{2}x^2y+9\right)\)
\(=\left(\frac{1}{2}x^2y+3\right)\cdot\left[\left(\frac{1}{2}x^2y\right)^2-\frac{1}{2}x^2y\cdot3+3^2\right]\)
\(=\left(\frac{1}{2}x^2y\right)^3+3^3\)
\(=\frac{1}{8}x^6y^3+27\)
Bài 2 :
Ta có : \(\left(\frac{1}{2}x^2y+3\right)\left(\frac{1}{4}x^4y^2-\frac{3}{2}x^2y+9\right)\)
\(=\frac{1}{8}x^6y^3+\frac{3}{4}x^4y^2-\frac{3}{4}x^4y^2-3x^2y+\frac{9}{2}x^2y+27\)
\(=\frac{1}{8}x^6y^3+\frac{3}{2}x^2y+27\)