$\sqrt[]{9x^2+6x+1}$ = $\sqrt[]{11-6\sqrt[]{2}}$ giải phương trình trên 20/07/2021 Bởi Kaylee $\sqrt[]{9x^2+6x+1}$ = $\sqrt[]{11-6\sqrt[]{2}}$ giải phương trình trên
Gửi bạn `\sqrt{9x^2+6x+1} = \sqrt{11-6\sqrt{2}}` `⇔ 9x^2 + 6x + 1 = 11 – 6\sqrt{2}` `⇔ 9x^2 + 6x + 6\sqrt{2} – 10 = 0` `⇔ x_{1,2} = (-6\pm\sqrt{6^2-4.9(6\sqrt{2}-10)))/(2.9)` `⇔ x_1 = (2-\sqrt{2})/3 , x_2 = (\sqrt{2}-4)/3` Vậy `S = {(2-\sqrt{2})/3 , (\sqrt{2}-4)/3}` Bình luận
`\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}` ĐK: `9x^2+6x+1>=0` với `AAx` `<=> \sqrt{(3x+1)^2}=\sqrt{2-2.3\sqrt{2}+9}` `<=> |3x+1|=\sqrt{(\sqrt{2}-3)^2}` `<=> |3x+1|=|\sqrt{2}-3|` `<=> |3x+1|=3-\sqrt{2}` `<=>`\(\left[ \begin{array}{l}3x+1=3-\sqrt{2}\\3x+1=\sqrt{2}-3\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}3x=2-\sqrt{2}\\3x=\sqrt{2}-4\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{2-\sqrt{2}}{3}\\x=\dfrac{\sqrt{2}-4}{3}\end{array} \right.\) Vậy `S={ (2-\sqrt{2})/3; (\sqrt{2}-4)/3 }` Bình luận
Gửi bạn
`\sqrt{9x^2+6x+1} = \sqrt{11-6\sqrt{2}}`
`⇔ 9x^2 + 6x + 1 = 11 – 6\sqrt{2}`
`⇔ 9x^2 + 6x + 6\sqrt{2} – 10 = 0`
`⇔ x_{1,2} = (-6\pm\sqrt{6^2-4.9(6\sqrt{2}-10)))/(2.9)`
`⇔ x_1 = (2-\sqrt{2})/3 , x_2 = (\sqrt{2}-4)/3`
Vậy `S = {(2-\sqrt{2})/3 , (\sqrt{2}-4)/3}`
`\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}` ĐK: `9x^2+6x+1>=0` với `AAx`
`<=> \sqrt{(3x+1)^2}=\sqrt{2-2.3\sqrt{2}+9}`
`<=> |3x+1|=\sqrt{(\sqrt{2}-3)^2}`
`<=> |3x+1|=|\sqrt{2}-3|`
`<=> |3x+1|=3-\sqrt{2}`
`<=>`\(\left[ \begin{array}{l}3x+1=3-\sqrt{2}\\3x+1=\sqrt{2}-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}3x=2-\sqrt{2}\\3x=\sqrt{2}-4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{2-\sqrt{2}}{3}\\x=\dfrac{\sqrt{2}-4}{3}\end{array} \right.\)
Vậy `S={ (2-\sqrt{2})/3; (\sqrt{2}-4)/3 }`