Toán 8 | Giải toán 8 | Giải toán lớp 8 | Giải bài tập toán 8 | Bài 2. Nhân đa thức với đa thức
Đề bài:
a) (\(\frac{1}{2}\)x + y)(\(\frac{1}{2}\)x + y);
b) (x – \(\frac{1}{2}\)y)(x – \(\frac{1}{2}\)y)
b) (x – \(\frac{1}{2}\)y)(x – \(\frac{1}{2}\)y) = x . x + x(-\(\frac{1}{2}\)y) + (-\(\frac{1}{2}\)y . x) + (-\(\frac{1}{2}\)y)(-\(\frac{1}{2}\)y) = x2 – \(\frac{1}{2}\)xy – \(\frac{1}{2}\)xy + \(\frac{1}{4}\) y2 = x2 – xy + \(\frac{1}{4}\) y2
Đề bài:
a) (\(\frac{1}{2}\)x + y)(\(\frac{1}{2}\)x + y);
b) (x – \(\frac{1}{2}\)y)(x – \(\frac{1}{2}\)y)
Lời giải toán
a) (\(\frac{1}{2}\)x + y)(\(\frac{1}{2}\)x + y) = \(\frac{1}{2}\)x . \(\frac{1}{2}\)x + \(\frac{1}{2}\)x . y + y . \(\frac{1}{2}\)x + y . y = \(\frac{1}{4}\)x2 + \(\frac{1}{2}\)xy + \(\frac{1}{2}\)xy + y2 = \(\frac{1}{4}\)x2 + xy + y2b) (x – \(\frac{1}{2}\)y)(x – \(\frac{1}{2}\)y) = x . x + x(-\(\frac{1}{2}\)y) + (-\(\frac{1}{2}\)y . x) + (-\(\frac{1}{2}\)y)(-\(\frac{1}{2}\)y) = x2 – \(\frac{1}{2}\)xy – \(\frac{1}{2}\)xy + \(\frac{1}{4}\) y2 = x2 – xy + \(\frac{1}{4}\) y2