Pitomath
Anonymous
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- asked 4 months agoVotes
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Giải SBT Toán 8 Bài 5: Phép cộng các phân thức đại số
a) 4x+2+ 2x−2+ 5x−64−x2;
b) 1−3x2x+3x−22x−1+ 3x−22x−4x2;
c) 1x2+ 6x+ 9+ 16x−x2−9+ xx2−9;
d) x2+2x3−1+ 2x2+x+ 1+11−x;
e) xx−2y+ xx+2y+ 4xy4y2−x2.
a)
4x+ 2+ 2x−2+ 5x−64−x2
=4x+ 2+ 2x−2− 5x−6x2−4
=4(x−2)+2(x+2)−(5x−6)(x−2).(x+2)
=4x−8+2x+ 4−5x+ 6(x−2).(x+2)
=x+ 2(x−2).(x+2)=1x−2
b) 1−3x2x+3x−22x−1+ 3x−22x−4x2
=1−3x2x+3x−22x−1− 3x−24x2−2x
=1−3x2x+3x−22x−1− 3x−22x(2x−1)
=(1−3x).(2x−1)+(3x−2).2x−(3x−2)2x(2x−1)
=2x−1− 6x2+3x+ 6x2−4x−3x+ 22x(2x−1)
=−2x+12x(2x−1)=−(2x−1)2x(2x−1)=−12x
c) 1x2+ 6x + 9+ 16x−x2−9+ xx2−9
d) x2+2x3−1+ 2x2+x+ 1+11−x
e) xx−2y+ xx+2y+ 4xy4y2−x2