Toán Lớp 8: (4x)/(x^2-4)+(x-2)/(2x+4)+(x+2)/(4-2x) ( x khác
±2)
a :rút gọn
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Toán Lớp 8: (4x)/(x^2-4)+(x-2)/(2x+4)+(x+2)/(4-2x) ( x khác
±2)
a :rút gọn
Comments ( 2 )
Giải đáp:
(4x)/(x^2-4)+(x-2)/(2x+4)+(x+2)/(4-2x) ( x \ne ±2)
= (4x)/[(x – 2).(x + 2)] + [x – 2]/[2.(x + 2)] + [x + 2]/[2.(2 – x)]
= (4x)/[(x – 2).(x + 2)] + [x – 2]/[2.(x + 2)] – [x + 2]/[2.(x – 2)]
= [8x]/[2.(x – 2).(x + 2)] + [(x – 2)^2]/[2.(x – 2).(x + 2)] – [(x + 2)^2]/[2.(x – 2).(x + 2)]
= [8x]/[2.(x – 2).(x + 2)] + [x^2 – 4x + 4]/[2.(x – 2).(x + 2)] – [x^2 + 4x + 4]/[2.(x – 2).(x + 2)]
= [8x + x^2 – 4x + 4 – x^2 – 4x – 4]/[2.(x – 2).(x + 2)]
= 0/[2.(x – 2).(x + 2)]
= 0
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Giải đáp:
Lời giải và giải thích chi tiết:
[4x]/[x^2 -4]+[x-2]/[2x+4]+[x+2]/[4-2x] (đk:x\ne -2;2)
=[4x]/[(x-2)(x+2)]+[x-2]/[2(x+2)]-[x+2]/[2(x-2)]
=[4x.2+(x-2)(x-2)-(x+2)(x+2)]/[2(x-2)(x+2)]
=[8x+x^2 -4x+4-x^2 -4x-4]/[2(x-2)(x+2)]
=0/[2(x-2)(x+2)]
=0