bài 1:Tìm x a, x^2-5x+1=0 b, 3x^2-12x-1=0 23/08/2021 Bởi Vivian bài 1:Tìm x a, x^2-5x+1=0 b, 3x^2-12x-1=0
Đáp án: Giải thích các bước giải: a) `x^2-5x+1=0` `⇔ x^2-5x+\frac{25}{4}-\frac{21}{4}=0` `⇔ (x-\frac{5}{2})^2-\frac{21}{4}=0` `⇔ (x-\frac{5}{2})^2-(\frac{\sqrt{21}}{2})^2=0` `⇔ (x-\frac{5}{2}+\frac{\sqrt{21}}{2})(x-\frac{5}{2}-\frac{\sqrt{21}}{2})=0` `⇔` \(\left[ \begin{array}{l}x-\dfrac{5}{2}+\dfrac{\sqrt{21}}{2}=0\\x-\dfrac{5}{2}-\dfrac{\sqrt{21}}{2}=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{5-\sqrt{21}}{2}\\x=\dfrac{5+\sqrt{21}}{2}\end{array} \right.\) Vậy `S={\frac{5-\sqrt{21}}{2};\frac{5+\sqrt{21}}{2}}` b) `3x^2-12x-1=0` `⇔ (\sqrt{3}x-2\sqrt{3})^2-13=0` `⇔ (\sqrt{3}x-2\sqrt{3})^2-(\sqrt{13})^2=0` `⇔ (\sqrt{3}x-2\sqrt{3}+\sqrt{13}).(\sqrt{3}x-2\sqrt{3}-\sqrt{13})=0` `⇔` \(\left[ \begin{array}{l}\sqrt{3}x-2\sqrt{3}+\sqrt{13}=0\\\sqrt{3}x-2\sqrt{3}-\sqrt{13}=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{6+\sqrt{39}}{3}\\x=\dfrac{6-\sqrt{39}}{3}\end{array} \right.\) Vậy `S={\frac{6+\sqrt{39}}{3};\frac{6-\sqrt{39}}{3}}` Bình luận
@Holliwood#: Đáp án: a) x1 = $\frac{5-√21}{2}$, x2 = $frac{5+√21}{2}$ b) x1 = $\frac{6-√39}{3}$, x2 = $frac{6+√39}{3}$ Giải thích các bước giải: a) x²-5x+1=0 ⇔ x = $\frac{-(-5)±√(-5)²-4.1.1}{2.1}$ ⇔ $\frac{5+√25-4}{2}$ ⇔ $\frac{5±√21}{2}$ ⇔ $\left \{ {{$\frac{5-√21}{2}$,} \atop {$\frac{5+√21}{2}$}} \right.$ ⇒ x1 = $\frac{5-√21}{2}$, x2 = $\frac{5+√21}{2}$ b) 3x²-12x-1=0 ⇔ x = $\frac{-(-12)±√(-12)²-4.3.(-1)}{2.3}$ ⇔ $\frac{12±√144+124}{6}$ ⇔ $\frac{12±√156}{6}$ ⇔ $\frac{12±2√39}{6}$ ⇔ $\left \{ {{$\frac{12-2√39}{6}$,} \atop {$\frac{12+2√39}{6}$}} \right.$ ⇔ $\left \{ {{$\frac{6-√39}{3}$,} \atop {$\frac{6+√39}{2}$}} \right.$ ⇒ x1 = $\frac{6-√39}{3}$, x2 = $\frac{6+√39}{3}$ Bn tham khaor nhes!!!!!!!!!!!!!!!!!!!!!!!!!!\ Xin ctlhn aj!!!!!!!!!!!!! Bình luận
Đáp án:
Giải thích các bước giải:
a) `x^2-5x+1=0`
`⇔ x^2-5x+\frac{25}{4}-\frac{21}{4}=0`
`⇔ (x-\frac{5}{2})^2-\frac{21}{4}=0`
`⇔ (x-\frac{5}{2})^2-(\frac{\sqrt{21}}{2})^2=0`
`⇔ (x-\frac{5}{2}+\frac{\sqrt{21}}{2})(x-\frac{5}{2}-\frac{\sqrt{21}}{2})=0`
`⇔` \(\left[ \begin{array}{l}x-\dfrac{5}{2}+\dfrac{\sqrt{21}}{2}=0\\x-\dfrac{5}{2}-\dfrac{\sqrt{21}}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{5-\sqrt{21}}{2}\\x=\dfrac{5+\sqrt{21}}{2}\end{array} \right.\)
Vậy `S={\frac{5-\sqrt{21}}{2};\frac{5+\sqrt{21}}{2}}`
b) `3x^2-12x-1=0`
`⇔ (\sqrt{3}x-2\sqrt{3})^2-13=0`
`⇔ (\sqrt{3}x-2\sqrt{3})^2-(\sqrt{13})^2=0`
`⇔ (\sqrt{3}x-2\sqrt{3}+\sqrt{13}).(\sqrt{3}x-2\sqrt{3}-\sqrt{13})=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{3}x-2\sqrt{3}+\sqrt{13}=0\\\sqrt{3}x-2\sqrt{3}-\sqrt{13}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{6+\sqrt{39}}{3}\\x=\dfrac{6-\sqrt{39}}{3}\end{array} \right.\)
Vậy `S={\frac{6+\sqrt{39}}{3};\frac{6-\sqrt{39}}{3}}`
@Holliwood#:
Đáp án:
a) x1 = $\frac{5-√21}{2}$, x2 = $frac{5+√21}{2}$
b) x1 = $\frac{6-√39}{3}$, x2 = $frac{6+√39}{3}$
Giải thích các bước giải:
a) x²-5x+1=0
⇔ x = $\frac{-(-5)±√(-5)²-4.1.1}{2.1}$
⇔ $\frac{5+√25-4}{2}$
⇔ $\frac{5±√21}{2}$
⇔ $\left \{ {{$\frac{5-√21}{2}$,} \atop {$\frac{5+√21}{2}$}} \right.$
⇒ x1 = $\frac{5-√21}{2}$, x2 = $\frac{5+√21}{2}$
b) 3x²-12x-1=0
⇔ x = $\frac{-(-12)±√(-12)²-4.3.(-1)}{2.3}$
⇔ $\frac{12±√144+124}{6}$
⇔ $\frac{12±√156}{6}$
⇔ $\frac{12±2√39}{6}$
⇔ $\left \{ {{$\frac{12-2√39}{6}$,} \atop {$\frac{12+2√39}{6}$}} \right.$
⇔ $\left \{ {{$\frac{6-√39}{3}$,} \atop {$\frac{6+√39}{2}$}} \right.$
⇒ x1 = $\frac{6-√39}{3}$, x2 = $\frac{6+√39}{3}$
Bn tham khaor nhes!!!!!!!!!!!!!!!!!!!!!!!!!!\
Xin ctlhn aj!!!!!!!!!!!!!