phép trừ các phân thức:a,1/x+2-x+1/x^3+8-1/2 07/08/2021 Bởi Lyla phép trừ các phân thức:a,1/x+2-x+1/x^3+8-1/2
Giải thích các bước giải: $\eqalign{ & \frac{{MD}}{{DC}}.\frac{{AH}}{{AM}}.\frac{{BC}}{{BH}} = 1 \cr & \Rightarrow \frac{{MD}}{{DC}}.2.2 = 1 \cr & \Rightarrow \frac{{MD}}{{DC}} = \frac{1}{4} \cr & \frac{{AD}}{{AB}} = \frac{{DK}}{{BH}} \cr & \frac{1}{{x + 2}} – x + \frac{1}{{{x^3} + 8}} – \frac{1}{2} \cr & = \frac{{{x^2} – 2x + 4}}{{(x + 2)({x^2} – 2x + 4)}} – \frac{{x({x^3} + 8)}}{{(x + 2)({x^2} – 2x + 4)}} + \frac{1}{{(x + 2)({x^2} – 2x + 4)}} – \frac{1}{2} \cr & = \frac{{{x^2} – 2x + 4 – {x^4} – 8x}}{{(x + 2)({x^2} – 2x + 4)}} – \frac{1}{2} \cr & = \frac{{2( – {x^4} + {x^2} – 10x + 4)}}{{2(x + 2)({x^2} – 2x + 4)}} – \frac{{{x^3} + 8}}{{2({x^3} + 8)}} \cr & = \frac{{ – 2{x^4} + 2{x^2} – 20x + 8 – {x^3} – 8}}{{2(x + 2)({x^2} – 2x + 4)}} \cr & = \frac{{ – 2{x^4} – {x^3} + 2{x^2} – 20x}}{{2(x + 2)({x^2} – 2x + 4)}} \cr} $ Bình luận
Giải thích các bước giải:
$\eqalign{ & \frac{{MD}}{{DC}}.\frac{{AH}}{{AM}}.\frac{{BC}}{{BH}} = 1 \cr & \Rightarrow \frac{{MD}}{{DC}}.2.2 = 1 \cr & \Rightarrow \frac{{MD}}{{DC}} = \frac{1}{4} \cr & \frac{{AD}}{{AB}} = \frac{{DK}}{{BH}} \cr & \frac{1}{{x + 2}} – x + \frac{1}{{{x^3} + 8}} – \frac{1}{2} \cr & = \frac{{{x^2} – 2x + 4}}{{(x + 2)({x^2} – 2x + 4)}} – \frac{{x({x^3} + 8)}}{{(x + 2)({x^2} – 2x + 4)}} + \frac{1}{{(x + 2)({x^2} – 2x + 4)}} – \frac{1}{2} \cr & = \frac{{{x^2} – 2x + 4 – {x^4} – 8x}}{{(x + 2)({x^2} – 2x + 4)}} – \frac{1}{2} \cr & = \frac{{2( – {x^4} + {x^2} – 10x + 4)}}{{2(x + 2)({x^2} – 2x + 4)}} – \frac{{{x^3} + 8}}{{2({x^3} + 8)}} \cr & = \frac{{ – 2{x^4} + 2{x^2} – 20x + 8 – {x^3} – 8}}{{2(x + 2)({x^2} – 2x + 4)}} \cr & = \frac{{ – 2{x^4} – {x^3} + 2{x^2} – 20x}}{{2(x + 2)({x^2} – 2x + 4)}} \cr} $