Giải pt sau $\frac{x}{x-3}$ -$\frac{x}{x-5}$= $\frac{x}{x-4}$ – $\frac{x}{x-6}$ 21/07/2021 Bởi Adeline Giải pt sau $\frac{x}{x-3}$ -$\frac{x}{x-5}$= $\frac{x}{x-4}$ – $\frac{x}{x-6}$
Đáp án: `S={0;9/2}` Giải thích các bước giải: `x/(x – 3) – x/(x – 5)=x/(x-4)-x/(x-6)(xne3,xne5,xne4,xne6)` $\\$ `<=> [x(x-5)-x(x-3)]/[(x-3)(x-5)]=[x(x-6)-x(x-4)]/[(x-4)(x-6)]` $\\$ `<=> (x^2-5x-x^2+3x)/[(x – 3)(x – 5)] = (x^2 – 6x – x^2 + 4x)/[(x-4)(x-6)]` $\\$ `<=> (-2x)/[(x – 3)(x – 5)] = (-2x)/[(x-4)(x-6)]` $\\$ `<=> (-2x)/[(x – 3)(x – 5)] – (-2x)/[(x – 4)(x – 6)]=0` $\\$ `<=> (-2x)(1/[(x-3)(x-5)] – 1/[(x-4)(x-6)]) = 0` TH1 : `-2x = 0<=>x=0(tm)` TH2 : `1/[(x-3)(x-5)] – 1/[(x-4)(x-6)]=0 ` $\\$ `<=> [(x-4)(x-6)-(x-3)(x-5)]/[(x-3)(x-5)(x-4)(x-6)] = 0` $\\$ `=> (x – 4)(x – 6) – (x – 3)(x – 5) = 0` $\\$ `<=> x^2 – 6x – 4x + 24 – (x^2 – 5x – 3x + 15) = 0` $\\$ `<=> x^2 – 6x – 4x + 24 – x^2 + 5x + 3x – 15 = 0` $\\$ `<=> -2x + 9 = 0 <=> -2x = -9 <=> x = 9/2(tm)` Vậy `S={0;9/2}` Bình luận
Đáp án: `S={0;9/2}`
Giải thích các bước giải:
`x/(x – 3) – x/(x – 5)=x/(x-4)-x/(x-6)(xne3,xne5,xne4,xne6)` $\\$ `<=> [x(x-5)-x(x-3)]/[(x-3)(x-5)]=[x(x-6)-x(x-4)]/[(x-4)(x-6)]` $\\$ `<=> (x^2-5x-x^2+3x)/[(x – 3)(x – 5)] = (x^2 – 6x – x^2 + 4x)/[(x-4)(x-6)]` $\\$ `<=> (-2x)/[(x – 3)(x – 5)] = (-2x)/[(x-4)(x-6)]` $\\$ `<=> (-2x)/[(x – 3)(x – 5)] – (-2x)/[(x – 4)(x – 6)]=0` $\\$ `<=> (-2x)(1/[(x-3)(x-5)] – 1/[(x-4)(x-6)]) = 0`
TH1 : `-2x = 0<=>x=0(tm)`
TH2 : `1/[(x-3)(x-5)] – 1/[(x-4)(x-6)]=0 ` $\\$ `<=> [(x-4)(x-6)-(x-3)(x-5)]/[(x-3)(x-5)(x-4)(x-6)] = 0` $\\$ `=> (x – 4)(x – 6) – (x – 3)(x – 5) = 0` $\\$ `<=> x^2 – 6x – 4x + 24 – (x^2 – 5x – 3x + 15) = 0` $\\$ `<=> x^2 – 6x – 4x + 24 – x^2 + 5x + 3x – 15 = 0` $\\$ `<=> -2x + 9 = 0 <=> -2x = -9 <=> x = 9/2(tm)`
Vậy `S={0;9/2}`