Giải các phương trình sau: a, x-30/10 +x-35/15 +x-47/27=-3 b, x^3-4x^2+x-4=0 25/11/2021 Bởi Hailey Giải các phương trình sau: a, x-30/10 +x-35/15 +x-47/27=-3 b, x^3-4x^2+x-4=0
a/ $\dfrac{x-30}{10}+\dfrac{x-35}{15}+\dfrac{x-47}{27}=-3$ $↔\dfrac{x-30}{10}+1+\dfrac{x-35}{15}+1+\dfrac{x-47}{27}+1=0$ $↔\dfrac{x-20}{10}+\dfrac{x-20}{15}+\dfrac{x-20}{27}=0$ $↔(x-20)(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{27})=0$ Vì $\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{27}\ne 0$ $↔x-20=0$ $↔x=20$ b/ $x³-4x²+x-4=0$ $↔x²(x-4)+(x-4)=0$ $↔(x²+1)(x-4)=0$ Vì $x²+1>0$ $↔x-4=0$ $↔x=4$ Bình luận
` a) ` ` \frac{x – 30}{10} + \frac{x – 35}{15} + \frac{x – 47}{27} = -3 ` ` \frac{x – 30}{10} + \frac{x – 35}{15} + \frac{x – 47}{27} + 3 = 0 ` ` => (1 + \frac{x – 30}{10}) + (1 + \frac{x – 35}{15}) + (1 + \frac{x – 47}{27}) = 0 ` ` => \frac{x – 20}{10} + \frac{x – 20}{15} + \frac{x – 20}{27} = 0 ` ` => (x – 20)(\frac{1}{10} + \frac{1}{15} + \frac{1}{27}) = 0 ` ` => x – 20 = 0 ` ` => x = 20 ` Vậy phương trình có 1 nghiệm duy nhất là: ` 20 ` ` b) ` ` x^3 – 4x^2 + x – 4 = 0 ` ` => x^{2}(x – 4) + (x – 4) = 0 ` ` => (x – 4)(x^{2} + 1) = 0 ` ` => ` \(\left[ \begin{array}{l}x-4=0\\x^2-1=0\end{array} \right.\) ` => x = 4 ` (Loại bỏ ` x^2 + 1 ` vì ` x^2 ≥ 0 `) Vậy phương trình có 1 nghiệm duy nhất là: ` 4 ` Bình luận
a/ $\dfrac{x-30}{10}+\dfrac{x-35}{15}+\dfrac{x-47}{27}=-3$
$↔\dfrac{x-30}{10}+1+\dfrac{x-35}{15}+1+\dfrac{x-47}{27}+1=0$
$↔\dfrac{x-20}{10}+\dfrac{x-20}{15}+\dfrac{x-20}{27}=0$
$↔(x-20)(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{27})=0$
Vì $\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{27}\ne 0$
$↔x-20=0$
$↔x=20$
b/ $x³-4x²+x-4=0$
$↔x²(x-4)+(x-4)=0$
$↔(x²+1)(x-4)=0$
Vì $x²+1>0$
$↔x-4=0$
$↔x=4$
` a) ` ` \frac{x – 30}{10} + \frac{x – 35}{15} + \frac{x – 47}{27} = -3 `
` \frac{x – 30}{10} + \frac{x – 35}{15} + \frac{x – 47}{27} + 3 = 0 `
` => (1 + \frac{x – 30}{10}) + (1 + \frac{x – 35}{15}) + (1 + \frac{x – 47}{27}) = 0 `
` => \frac{x – 20}{10} + \frac{x – 20}{15} + \frac{x – 20}{27} = 0 `
` => (x – 20)(\frac{1}{10} + \frac{1}{15} + \frac{1}{27}) = 0 `
` => x – 20 = 0 `
` => x = 20 `
Vậy phương trình có 1 nghiệm duy nhất là: ` 20 `
` b) ` ` x^3 – 4x^2 + x – 4 = 0 `
` => x^{2}(x – 4) + (x – 4) = 0 `
` => (x – 4)(x^{2} + 1) = 0 `
` => ` \(\left[ \begin{array}{l}x-4=0\\x^2-1=0\end{array} \right.\)
` => x = 4 ` (Loại bỏ ` x^2 + 1 ` vì ` x^2 ≥ 0 `)
Vậy phương trình có 1 nghiệm duy nhất là: ` 4 `