Toán Lớp 8: Tính giá trị biểu thức: a) M = (a – 2b)(a^2+ 2ab + 4b^2) + (2b − a)^3tại a = -1; b = 2; b) N = (2xy – 2)(2xy + 3) – (1 − 2xy)^2 tại x

\(\begin{array}{l}
a)M = (a – 2b)({a^2} + 2ab + 4{b^2}) + {(2b – a)^3}\\
= {a^3} – 8{b^3} + 8{b^3} – 12{b^2}a + 6b{a^2} – {a^3}\\
= 6{{\rm{a}}^2}b – 12{\rm{a}}{b^2}\\
thay:a = – 1;b = 2;\\
M = 6{\left( { – 1} \right)^2}.2 – 12\left( { – 1} \right),{2^2}\\
= 12 + 48\\
= 60\\
vay:M = 60\\
b)N = (2xy – 2)(2xy + 3) – {(1 – 2xy)^2}\\
= 4{{\rm{x}}^2}{y^2} + 6{\rm{x}}y – 4{\rm{x}}y – 6 – \left( {1 – 4{\rm{x}}y + 4{{\rm{x}}^2}{y^2}} \right)\\
= 4{{\rm{x}}^2}{y^2} + 2{\rm{x}}y – 6 – 1 + 4{\rm{x}}y – 4{{\rm{x}}^2}{y^2}\\
= 6{\rm{x}}y – 7\\
Thay:x = \frac{1}{2};y = – 1.\\
N = 6.\frac{1}{2}\left( { – 1} \right) – 7\\
= – 3 – 7\\
= – 10\\
vay:n = – 10
\end{array}\)