Giải bài 35 trang 50 - Sách giáo khoa Toán 8 tập 1
Đề bài
Thực hiện các phép tính:
Hướng dẫn giải
a) Ta có 9−x2=−(x2−9)=−(x−3)(x+3)
MTC = (x−3)(x+3)
Do đó : x+1x–3–1–xx+3–2x(1–x)9–x2
=x+1x–3–1–xx+3+2x(1–x)(x−3)(x+3)
=(x+1)(x+3)(x–3)(x+3)–(1–x)(x–3)(x+3)(x–3)+2x(1–x)(x−3)(x+3)
=(x+1)(x+3)+(x−1)(x–3)+2x(1–x)(x–3)(x+3)
=x2+4x+3+x2−4x+3+2x–2x2(x–3)(x+3)
=2x+6(x–3)(x+3)=2(x+3)(x–3)(x+3)=2x–3.
b) Ta có : 1−x2=(1−x)(1+x)
MTC = (1−x)2(x+1)
Do đó : 3x+1(x–1)2–1x+1+x+31–x2
=3x+1(x–1)2–1x+1–x+3(1−x)(1+x)
=(3x+1)(x+1)(x–1)2(x+1)–(x–1)2(x–1)2(x+1)–(x+3)(x–1)(x–1)2(x+1)
=3x2+4x+1–x2+2x–1–x2–2x+3(x–1)2(x+1)
=x2+4x+3(x–1)2(x+1)=(x+1)(x+3)(x–1)2(x+1)=x+3(x–1)2.