Giải phương trình : `x^2 + x + 6 + 2x\sqrt{x+3} = 4 ( x + \sqrt{x + 3 })` 11/07/2021 Bởi Liliana Giải phương trình : `x^2 + x + 6 + 2x\sqrt{x+3} = 4 ( x + \sqrt{x + 3 })`
Đáp án: Giải thích các bước giải: `x^2 + x + 6 + 2x\sqrt{x+3} = 4 ( x + \sqrt{x + 3 })` ĐK: `x \ge -3` `⇔ x^2-3x+2x\sqrt{x+3}-4\sqrt{x+3}+6=0` `⇔ (x+\sqrt{x+3})^2-4x-4\sqrt{x+3}+3=0` `⇔ (x+\sqrt{x+3})^2-4(x+\sqrt{x+3})+3=0` Đặt `x+\sqrt{x+3}=t` ta có: `t^2-4t+3=0` `⇔ (t-1)(t-3)=0` `⇔` \(\left[ \begin{array}{l}t=1\\t=3\end{array} \right.\) `+) t=1⇒x+\sqrt{x+3}=1` `⇔ \sqrt{x+3}=1-x` ĐK: `x \le 1` `⇔ x+3=1-2x+x^2` `⇔ x^2-3x-2=0` `⇔` \(\left[ \begin{array}{l}x=\dfrac{3+\sqrt{17}}{2}\ (L)\\x=\dfrac{3-\sqrt{17}}{2}\ (TM)\end{array} \right.\) `+) t=3⇒ x+\sqrt{x+3}=3` `⇔ \sqrt{x+3}=3-x` ĐK: `x \le 3 `⇔ x+3=9-6x+x^2` `⇔ x^2-7x+6=0` `⇔ (x-1)(x-6)=0` `⇔` `⇔` \(\left[ \begin{array}{l}x=1\ (TM)\\x=6\ (L)\end{array} \right.\) Vậy `S={\frac{3-\sqrt{17}}{2};1}` Bình luận
`ĐKXĐ: x\ge -3` `x^2+x+6+2x\sqrt(x+3)=4(x+\sqrt(x+3))` `⇔[x^2+2x\sqrt(x+3)+(x+3)]+3=4(x+\sqrt(x+3))` `⇔(x+\sqrt(x+3))^2-4(x+\sqrt(x+3))+3=0` Đặt `x+\sqrt(x+3)=y` `Pt⇔y^2-4y+3=0` `⇔(y-1)(y-3)=0` \(⇔\left[ \begin{array}{l}y=1\\y=3\end{array} \right.⇔\left[ \begin{array}{l}x+\sqrt{x+3}=1\\x+\sqrt{x+3}=3\end{array} \right.⇔\left[ \begin{array}{l}\sqrt{x+3}=1-x(x\leq 1)\\\sqrt{x+3}=3-x(x\leq 3)\end{array} \right.⇔\left[ \begin{array}{l}x+3=x^2-2x+1\\x+3=x^2-6x+9\end{array} \right.⇔\left[ \begin{array}{l}x^2-3x-2=0\\x^2-7x+6=0\end{array} \right.⇔\left[ \begin{array}{l}x=\dfrac{3-\sqrt{17}}{2}\\x=1\end{array} \right.\) Vậy `S={1; (3-\sqrt17)/2}` Bình luận
Đáp án:
Giải thích các bước giải:
`x^2 + x + 6 + 2x\sqrt{x+3} = 4 ( x + \sqrt{x + 3 })`
ĐK: `x \ge -3`
`⇔ x^2-3x+2x\sqrt{x+3}-4\sqrt{x+3}+6=0`
`⇔ (x+\sqrt{x+3})^2-4x-4\sqrt{x+3}+3=0`
`⇔ (x+\sqrt{x+3})^2-4(x+\sqrt{x+3})+3=0`
Đặt `x+\sqrt{x+3}=t` ta có:
`t^2-4t+3=0`
`⇔ (t-1)(t-3)=0`
`⇔` \(\left[ \begin{array}{l}t=1\\t=3\end{array} \right.\)
`+) t=1⇒x+\sqrt{x+3}=1`
`⇔ \sqrt{x+3}=1-x`
ĐK: `x \le 1`
`⇔ x+3=1-2x+x^2`
`⇔ x^2-3x-2=0`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{3+\sqrt{17}}{2}\ (L)\\x=\dfrac{3-\sqrt{17}}{2}\ (TM)\end{array} \right.\)
`+) t=3⇒ x+\sqrt{x+3}=3`
`⇔ \sqrt{x+3}=3-x`
ĐK: `x \le 3
`⇔ x+3=9-6x+x^2`
`⇔ x^2-7x+6=0`
`⇔ (x-1)(x-6)=0`
`⇔` `⇔` \(\left[ \begin{array}{l}x=1\ (TM)\\x=6\ (L)\end{array} \right.\)
Vậy `S={\frac{3-\sqrt{17}}{2};1}`
`ĐKXĐ: x\ge -3`
`x^2+x+6+2x\sqrt(x+3)=4(x+\sqrt(x+3))`
`⇔[x^2+2x\sqrt(x+3)+(x+3)]+3=4(x+\sqrt(x+3))`
`⇔(x+\sqrt(x+3))^2-4(x+\sqrt(x+3))+3=0`
Đặt `x+\sqrt(x+3)=y`
`Pt⇔y^2-4y+3=0`
`⇔(y-1)(y-3)=0`
\(⇔\left[ \begin{array}{l}y=1\\y=3\end{array} \right.⇔\left[ \begin{array}{l}x+\sqrt{x+3}=1\\x+\sqrt{x+3}=3\end{array} \right.⇔\left[ \begin{array}{l}\sqrt{x+3}=1-x(x\leq 1)\\\sqrt{x+3}=3-x(x\leq 3)\end{array} \right.⇔\left[ \begin{array}{l}x+3=x^2-2x+1\\x+3=x^2-6x+9\end{array} \right.⇔\left[ \begin{array}{l}x^2-3x-2=0\\x^2-7x+6=0\end{array} \right.⇔\left[ \begin{array}{l}x=\dfrac{3-\sqrt{17}}{2}\\x=1\end{array} \right.\)
Vậy `S={1; (3-\sqrt17)/2}`