Toán x + 3$\sqrt{x+2}$ = 0 x + $\sqrt{x – 1}$ = 13 23/07/2021 By Caroline x + 3$\sqrt{x+2}$ = 0 x + $\sqrt{x – 1}$ = 13
Đáp án: Giải thích các bước giải: `x+3\sqrt{x+2}=0` `ĐK: x \ge -2` `⇔ x+2-2+3\sqrt{x+2}=0` Đặt `\sqrt{x+2}=t\ (t \ge 0)` `⇔ t^2+3t-2=0` `⇔` \(\left[ \begin{array}{l}t_1=\dfrac{-3+\sqrt{17}}{2}\ (TM)\\x=-\dfrac{-3-\sqrt{17}}{2}\ (Loại)\end{array} \right.\) `t=\frac{-3+\sqrt{17}}{2}⇔ \sqrt{x+2}=\frac{-3+\sqrt{17}}{2}` `⇔ x+2=\frac{13-3\sqrt{17}}{2}` `⇔ x=\frac{9-3\sqrt{17}}{2}\ (TM)` Vậy `S={\frac{9-3\sqrt{17}}{2}}` b) `x+\sqrt{x-1}=13` `ĐK: x \ge 1` `⇔ x-13+\sqrt{x-1}=0` Đặt `\sqrt{x-1}=t\ (t \ge 0)` `⇔ t^2+t-12=0` `⇔` \(\left[ \begin{array}{l}t_1=3\ (TM)\\t_2=-4\ (Loại)\end{array} \right.\) `t=3⇔ \sqrt{x-1}=3` `⇔ x-1=9` `⇔ x=10\ (TM)` Vậy `S={10}` Trả lời
` a) ` ` x + 3\sqrt{x + 2} = 0 ` ` <=> 3\sqrt{x + 2} = -x ` ` <=> 9(x + 2) = x^2 ` ` <=> 9x + 18 – x^2 = 0 ` ` <=> -x^2 + 9x + 18 = 0 ` ` <=> x^2 – 9x – 18 = 0 ` ` <=> x = \frac{-(-9) ± \sqrt{(-9)^{2} – 4 . 1 . (-18)}}{2.1} ` ` <=> x = \frac{9±\sqrt{153}}{2} ` ` <=> x = \frac{9±3\sqrt{17}}{2} ` * Nếu ` x = \frac{9 + 3\sqrt{17}}{2} ` ` <=> \frac{9 + 3\sqrt{17}}{2} + 3\sqrt{\frac{9+3\sqrt{17}}{2} + 2} = 0 ` ` <=> 21,36932 = 0 ` `(vô` `lí)` * Nếu ` x = \frac{9 – 3\sqrt{17}}{2} ` ` <=> \frac{9 + 3\sqrt{17}}{2} + 3\sqrt{\frac{9+3\sqrt{17}}{2} + 2} = 0 ` ` <=> 0 = 0 ` `(tm)` ` => x = \frac{9 – 3\sqrt{17}}{2} ` ` b) ` ` x + \sqrt{x – 1} = 13 ` ` <=> \sqrt{x – 1} = 13 – x ` ` <=> x – 1 = 169 – 26x + x^{2} ` ` <=> -x^{2} + 27x – 170 = 0 ` ` <=> x^{2} – 27x + 170 = 0 ` ` <=> x^{2} – 10x – 17x + 170 = 0 ` ` <=> (x – 10)(x – 17) = 0 ` ` <=> ` \(\left[ \begin{array}{l}x=10\\x=17\end{array} \right.\) * Nếu ` x = 10 ` ` <=> 10 + \sqrt{10 – 1} = 13 ` ` <=> 13 = 13 ` `(tm) ` * Nếu ` x = 17 ` ` <=> 17 + \sqrt{17 – 1} = 13 ` ` <=> 21 = 13 ` `(vô` `lí)` ` => x = 10 ` Trả lời
Đáp án:
Giải thích các bước giải:
`x+3\sqrt{x+2}=0`
`ĐK: x \ge -2`
`⇔ x+2-2+3\sqrt{x+2}=0`
Đặt `\sqrt{x+2}=t\ (t \ge 0)`
`⇔ t^2+3t-2=0`
`⇔` \(\left[ \begin{array}{l}t_1=\dfrac{-3+\sqrt{17}}{2}\ (TM)\\x=-\dfrac{-3-\sqrt{17}}{2}\ (Loại)\end{array} \right.\)
`t=\frac{-3+\sqrt{17}}{2}⇔ \sqrt{x+2}=\frac{-3+\sqrt{17}}{2}`
`⇔ x+2=\frac{13-3\sqrt{17}}{2}`
`⇔ x=\frac{9-3\sqrt{17}}{2}\ (TM)`
Vậy `S={\frac{9-3\sqrt{17}}{2}}`
b) `x+\sqrt{x-1}=13`
`ĐK: x \ge 1`
`⇔ x-13+\sqrt{x-1}=0`
Đặt `\sqrt{x-1}=t\ (t \ge 0)`
`⇔ t^2+t-12=0`
`⇔` \(\left[ \begin{array}{l}t_1=3\ (TM)\\t_2=-4\ (Loại)\end{array} \right.\)
`t=3⇔ \sqrt{x-1}=3`
`⇔ x-1=9`
`⇔ x=10\ (TM)`
Vậy `S={10}`
` a) ` ` x + 3\sqrt{x + 2} = 0 `
` <=> 3\sqrt{x + 2} = -x `
` <=> 9(x + 2) = x^2 `
` <=> 9x + 18 – x^2 = 0 `
` <=> -x^2 + 9x + 18 = 0 `
` <=> x^2 – 9x – 18 = 0 `
` <=> x = \frac{-(-9) ± \sqrt{(-9)^{2} – 4 . 1 . (-18)}}{2.1} `
` <=> x = \frac{9±\sqrt{153}}{2} `
` <=> x = \frac{9±3\sqrt{17}}{2} `
* Nếu ` x = \frac{9 + 3\sqrt{17}}{2} `
` <=> \frac{9 + 3\sqrt{17}}{2} + 3\sqrt{\frac{9+3\sqrt{17}}{2} + 2} = 0 `
` <=> 21,36932 = 0 ` `(vô` `lí)`
* Nếu ` x = \frac{9 – 3\sqrt{17}}{2} `
` <=> \frac{9 + 3\sqrt{17}}{2} + 3\sqrt{\frac{9+3\sqrt{17}}{2} + 2} = 0 `
` <=> 0 = 0 ` `(tm)`
` => x = \frac{9 – 3\sqrt{17}}{2} `
` b) ` ` x + \sqrt{x – 1} = 13 `
` <=> \sqrt{x – 1} = 13 – x `
` <=> x – 1 = 169 – 26x + x^{2} `
` <=> -x^{2} + 27x – 170 = 0 `
` <=> x^{2} – 27x + 170 = 0 `
` <=> x^{2} – 10x – 17x + 170 = 0 `
` <=> (x – 10)(x – 17) = 0 `
` <=> ` \(\left[ \begin{array}{l}x=10\\x=17\end{array} \right.\)
* Nếu ` x = 10 `
` <=> 10 + \sqrt{10 – 1} = 13 `
` <=> 13 = 13 ` `(tm) `
* Nếu ` x = 17 `
` <=> 17 + \sqrt{17 – 1} = 13 `
` <=> 21 = 13 ` `(vô` `lí)`
` => x = 10 `