Giải bài 7 trang 10 sách bài tập toán 8 – Chân trời sáng tạo

Đề bài

Tính:

a) \(3a\left( {a – b} \right) – b\left( {b – 3a} \right)\);

b) \(3{a^2}\left( {2a + b} \right) – 2b\left( {4{a^2} – b} \right)\);

c) \(\left( {a + b} \right)\left( {a – b} \right) – \left( {a – 1} \right)\left( {a – 2} \right)\);

d) \(b\left( {3{b^2} – {a^3}} \right) + \left( {{a^2} + 3b} \right)\left( {ab – {b^2}} \right)\).

Lời giải chi tiết

a) \(3a\left( {a – b} \right) – b\left( {b – 3a} \right) = 3{a^2} – 3ab – {b^2} + 3ab = 3{a^2} + \left( {3ab – 3ab} \right) – {b^2} = 3{a^2} – {b^2}\);

b) \(3{a^2}\left( {2a + b} \right) – 2b\left( {4{a^2} – b} \right) = 6{a^3} + 3{a^2}b – 8{a^2}b + 2{b^2}\)

\( = 6{a^3} + \left( {3{a^2}b – 8{a^2}b} \right) + 2{b^2} = 6{a^3} – 5{a^2}b + 2{b^2}\)

c) \(\left( {a + b} \right)\left( {a – b} \right) – \left( {a – 1} \right)\left( {a – 2} \right) = a\left( {a – b} \right) + b\left( {a – b} \right) – a\left( {a – 2} \right) + \left( {a – 2} \right)\)

\( = {a^2} – ab + ab – {b^2} – {a^2} + 2a + a – 2 = \left( {{a^2} – {a^2}} \right) + \left( {ab – ab} \right) – {b^2} + \left( {2a + a} \right) – 2\)

\( =  – {b^2} + 3a – 2\)

d) \(b\left( {3{b^2} – {a^3}} \right) + \left( {{a^2} + 3b} \right)\left( {ab – {b^2}} \right) = 3{b^3} – {a^3}b + {a^2}\left( {ab – {b^2}} \right) + 3b\left( {ab – {b^2}} \right)\)

\( = 3{b^3} – {a^3}b + {a^3}b – {a^2}{b^2} + 3a{b^2} – 3{b^3} = \left( {3{b^3} – 3{b^3}} \right) + \left( {{a^3}b – {a^3}b} \right) – {a^2}{b^2} + 3a{b^2}\)

\( =  – {a^2}{b^2} + 3a{b^2}\)