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Toán Lớp 8: Tìm x 1) [2x – 1]/[x^2 + 4x – 5] + [x-2]/[x^2 – 10x +9] = 3x – 12/x^4 – 4x – 45

Home/ Toán Lớp 8: Tìm x 1) [2x – 1]/[x^2 + 4x – 5] + [x-2]/[x^2 – 10x +9] = 3x – 12/x^4 – 4x – 45

Toán Lớp 8: Tìm x 1) [2x – 1]/[x^2 + 4x – 5] + [x-2]/[x^2 – 10x +9] = 3x – 12/x^4 – 4x – 45

Madelyn Tháng Tám 21, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Tìm x
1) [2x – 1]/[x^2 + 4x – 5] + [x-2]/[x^2 – 10x +9] = 3x – 12/x^4 – 4x – 45

Comments ( 1 )

catlantuong
21 Tháng Tám, 2022 at 10:43 chiều

Reply

~ gửi bạn ~

Giải đáp:

x = -18

Lời giải và giải thích chi tiết:

1)

[2x – 1]/[x^2 + 4x – 5] + [x-2]/[x^2 – 10x +9] = [3x – 12]/[x^2 – 4x – 45]

<=> [2x – 1]/[x^2 -x + 5x – 5] + [x-2]/[x^2 -x – 9x +9] – [3x – 12]/[x^2 + 5x – 9x – 45] = 0

<=> [2x – 1]/[(x – 1).(x + 5)] + [x-2]/[(x – 1).(x – 9)] – [3x – 12]/[(x + 5).(x – 9)] = 0

=> (2x−1).(x−9)+(x−2).(x+5)−(3x−12).(x−1)=0

⇔2x^2−19x+9+x^2+3x−15−3x^2+15x−12=0

<=> (2x^2 + x^2 – 3x^2) + (-19x+ 3x + 15x) + (9 – 15 -12) = 0

<=> -x – 18 = 0

<=> -x = 18

<=> x = -18

Vậy x = -18

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