Toán Giai pt sau: 5x/x^2-x-6 + x/3-x = 2/x+2 – 1 16/07/2021 By Vivian Giai pt sau: 5x/x^2-x-6 + x/3-x = 2/x+2 – 1
`(5x)/(x^2-x-6) + x/(3-x) = 2/(x+2) – 1` ` (x\ne-2; x\ne3)` `<=>(5x)/(x^2+2x-3x-6) + x/(3-x) = 2/(x+2) – 1` `<=>(5x)/(x(x+2)-3(x+2)) + x/(3-x) = 2/(x+2) – 1` `<=>(5x)/((x-3)(x+2))-x/(x-3)=2/(x+2)-1` `<=>(5x)/((x-3)(x+2))-(x(x+2))/((x-3)(x+2))=(2(x-3))/((x-3)(x+2))-((x-3)(x+2))/((x-3)(x+2))` `<=>(5x-x(x+2))/((x-3)(x+2))=(2(x-3)-(x-3)(x+2))/((x-3)(x+2))` `=>5x-x(x+2)=2(x-3)-(x-3)(x+2)` `<=>5x-x^2-2x=2x-6-x^2-2x+3x+6` `<=> 5x -2x -2x+2x-x^2+x^2-3x=-6+6` `<=>0x=0` (Luôn đúng với `∀x∈R`) Vậy: `S∈R` ` (x\ne-2; x\ne3)` Trả lời
Đáp án+Giải thích các bước giải: `(5x)/(x^2-x-6) + x/(3-x) = 2/(x+2) – 1 (ĐKXĐ: x\ne-2; x\ne3)` `<=> (5x)/((x-3)(x+2))-x/(x-3)=2/(x+2)-1` `<=> (5x)/((x-3)(x+2))-(x(x+2))/((x-3)(x+2))=(2(x-3))/((x-3)(x+2))-((x-3)(x+2))/((x-3)(x+2))` `=> 5x – x^2 -2x = 2x – 6 – x^2 -2x+3x+6` `<=> 5x -2x = 2x – 6 -2x+3x+6` `<=> 3x = 3x ` `<=> x ∈ RR` Vậy `x ∈ RR` trừ `-2` và `3` Trả lời