gpt a) 3x^2-x-1=0 b) x^2+2x-2=0 c) x^2-4x-6=0 28/09/2021 Bởi Madelyn gpt a) 3x^2-x-1=0 b) x^2+2x-2=0 c) x^2-4x-6=0
$a)$ $ 3x^2 -x -1 = 0 \to 3(x^2 – \dfrac{1}{3}x – \dfrac{1}{3}) = 0$ $ \to x^2 – \dfrac{1}{3}x – \dfrac{1}{3} = 0$ $ \to x^2 -\dfrac{1- \sqrt{13}}{6}x – \dfrac{1+ \sqrt{13}}{6}x – \dfrac{1- \sqrt{13}}{6}. \dfrac{1+ \sqrt{13}}{6}$ $\to x(x – \dfrac{1- \sqrt{13}}{6}) – \dfrac{1+ \sqrt{13}}{6}( x – \dfrac{1- \sqrt{13}}{6})$ $ \to (x – \dfrac{1- \sqrt{13}}{6}) ( x – \dfrac{1+ \sqrt{13}}{6}) = 0$ $ \to$ \(\left[ \begin{array}{l}x= \dfrac{1- \sqrt{13}}{6}\\x=\dfrac{1+ \sqrt{13}}{6}\end{array} \right.\) $b)$ $ x^2 + 2x -2 = 0 $ $\to x^2 + (\sqrt{3}+1)x – (\sqrt{3}-1)x – (\sqrt{3}+1)(\sqrt{3}-1) =0$ $ \to x(x + \sqrt{3}+1) – (\sqrt{3}-1)(x+ \sqrt{3}+1) = 0$ $ \to [x – (\sqrt{3}-1)](x + \sqrt{3}+1) = 0$ $ \to$ \(\left[ \begin{array}{l}x= \sqrt{3}-1\\x=-1 -\sqrt{3} \end{array} \right.\) $c)$ $ x^2- 4x -6 = 0$ $\to x^2 – (2-\sqrt{10})x – (2+ \sqrt{10}x + (2-\sqrt{10})(2+ \sqrt{10}) = 0$ $\to x[x – (2-\sqrt{10})] – (2+ \sqrt{10})[x – (2-\sqrt{10})] = 0$ $\to[x- (2+ \sqrt{10})][x – (2-\sqrt{10})] = 0$ $ \to$ \(\left[ \begin{array}{l}x=2+ \sqrt{10} \\x=2-\sqrt{10}\end{array} \right.\) Bình luận
Đáp án+Giải thích các bước giải: `a)3x^2-x-1=0` `<=>3(x^2-1/3x-1/3)=0` `<=>x^2-1/3x-1/3=0` `<=>x^2-2.x.\frac{1}{6}+(1/6)^2-13/36`=0` `<=>(x-1/6)^2-(\frac{\sqrt{13}}{6})^2` `<=>(x-1/6-\frac{\sqrt{13}}{6})(x-1/6+\frac{\sqrt{13}}{6})=0` `TH1:` `x-1/6-\frac{\sqrt{13}}{6}=0` `<=>x-1/6=0+\frac{\sqrt{13}}{6}` `<=>x=\frac{\sqrt{13}}{6}+1/6` `<=>x=\frac{\sqrt{13}+1}{6}` `TH2:` `x-1/6+\frac{\sqrt{13}}{6}=0` `<=>x-1/6=0-\frac{\sqrt{13}}{6}` `<=>x=-\frac{\sqrt{13}}{6}+1/6` `<=>x=\frac{1-\sqrt{13}}{6}` Vậy `S=\{\frac{\sqrt{13}+1}{6};\frac{1-\sqrt{13}}{6}\}` `b)x^2+2x-2=0` `<=>x^2+2x+1-3=0` `<=>x^2+2.x.1+1^2-3=0` `<=>(x-1)^2-(\sqrt{3})^2=0` `<=>(x-1-\sqrt{3})(x-1+\sqrt{3})=0` `TH1:` `x-1-\sqrt{3}=0` `<=>x-1=0+\sqrt{3}` `<=>x=\sqrt{3}+1` `TH2:` `x-1+\sqrt{3}=0` `<=>x-1=0-\sqrt{3}` `<=>x=-\sqrt{3}+1` `<=>x=1-\sqrt{3}` Vậy `S=\{\sqrt{3}+1;1-\sqrt{3}\}` `c)x^2-4x-6=0` `<=>x^2-4x+4-10=0` `<=>(x-2)^2-(\sqrt{10})^2=0` `<=>(x-2-\sqrt{10})(x-2+\sqrt{10})=0` `TH1:x-2-\sqrt{10}=0` `<=>x-2=0+\sqrt{10}` `<=>x=\sqrt{10}+2` `TH2:x-2+\sqrt{10}=0` `<=>x-2=0-\sqrt{10}` `<=>x=-\sqrt{10}+2` `<=>x=2-\sqrt{10}` Vậy `S=\{\sqrt{10}+2;2-\sqrt{10}\}` Bình luận
$a)$
$ 3x^2 -x -1 = 0 \to 3(x^2 – \dfrac{1}{3}x – \dfrac{1}{3}) = 0$
$ \to x^2 – \dfrac{1}{3}x – \dfrac{1}{3} = 0$
$ \to x^2 -\dfrac{1- \sqrt{13}}{6}x – \dfrac{1+ \sqrt{13}}{6}x – \dfrac{1- \sqrt{13}}{6}. \dfrac{1+ \sqrt{13}}{6}$
$\to x(x – \dfrac{1- \sqrt{13}}{6}) – \dfrac{1+ \sqrt{13}}{6}( x – \dfrac{1- \sqrt{13}}{6})$
$ \to (x – \dfrac{1- \sqrt{13}}{6}) ( x – \dfrac{1+ \sqrt{13}}{6}) = 0$
$ \to$ \(\left[ \begin{array}{l}x= \dfrac{1- \sqrt{13}}{6}\\x=\dfrac{1+ \sqrt{13}}{6}\end{array} \right.\)
$b)$
$ x^2 + 2x -2 = 0 $
$\to x^2 + (\sqrt{3}+1)x – (\sqrt{3}-1)x – (\sqrt{3}+1)(\sqrt{3}-1) =0$
$ \to x(x + \sqrt{3}+1) – (\sqrt{3}-1)(x+ \sqrt{3}+1) = 0$
$ \to [x – (\sqrt{3}-1)](x + \sqrt{3}+1) = 0$
$ \to$ \(\left[ \begin{array}{l}x= \sqrt{3}-1\\x=-1 -\sqrt{3} \end{array} \right.\)
$c)$
$ x^2- 4x -6 = 0$
$\to x^2 – (2-\sqrt{10})x – (2+ \sqrt{10}x + (2-\sqrt{10})(2+ \sqrt{10}) = 0$
$\to x[x – (2-\sqrt{10})] – (2+ \sqrt{10})[x – (2-\sqrt{10})] = 0$
$\to[x- (2+ \sqrt{10})][x – (2-\sqrt{10})] = 0$
$ \to$ \(\left[ \begin{array}{l}x=2+ \sqrt{10} \\x=2-\sqrt{10}\end{array} \right.\)
Đáp án+Giải thích các bước giải:
`a)3x^2-x-1=0`
`<=>3(x^2-1/3x-1/3)=0`
`<=>x^2-1/3x-1/3=0`
`<=>x^2-2.x.\frac{1}{6}+(1/6)^2-13/36`=0`
`<=>(x-1/6)^2-(\frac{\sqrt{13}}{6})^2`
`<=>(x-1/6-\frac{\sqrt{13}}{6})(x-1/6+\frac{\sqrt{13}}{6})=0`
`TH1:`
`x-1/6-\frac{\sqrt{13}}{6}=0`
`<=>x-1/6=0+\frac{\sqrt{13}}{6}`
`<=>x=\frac{\sqrt{13}}{6}+1/6`
`<=>x=\frac{\sqrt{13}+1}{6}`
`TH2:`
`x-1/6+\frac{\sqrt{13}}{6}=0`
`<=>x-1/6=0-\frac{\sqrt{13}}{6}`
`<=>x=-\frac{\sqrt{13}}{6}+1/6`
`<=>x=\frac{1-\sqrt{13}}{6}`
Vậy `S=\{\frac{\sqrt{13}+1}{6};\frac{1-\sqrt{13}}{6}\}`
`b)x^2+2x-2=0`
`<=>x^2+2x+1-3=0`
`<=>x^2+2.x.1+1^2-3=0`
`<=>(x-1)^2-(\sqrt{3})^2=0`
`<=>(x-1-\sqrt{3})(x-1+\sqrt{3})=0`
`TH1:`
`x-1-\sqrt{3}=0`
`<=>x-1=0+\sqrt{3}`
`<=>x=\sqrt{3}+1`
`TH2:`
`x-1+\sqrt{3}=0`
`<=>x-1=0-\sqrt{3}`
`<=>x=-\sqrt{3}+1`
`<=>x=1-\sqrt{3}`
Vậy `S=\{\sqrt{3}+1;1-\sqrt{3}\}`
`c)x^2-4x-6=0`
`<=>x^2-4x+4-10=0`
`<=>(x-2)^2-(\sqrt{10})^2=0`
`<=>(x-2-\sqrt{10})(x-2+\sqrt{10})=0`
`TH1:x-2-\sqrt{10}=0`
`<=>x-2=0+\sqrt{10}`
`<=>x=\sqrt{10}+2`
`TH2:x-2+\sqrt{10}=0`
`<=>x-2=0-\sqrt{10}`
`<=>x=-\sqrt{10}+2`
`<=>x=2-\sqrt{10}`
Vậy `S=\{\sqrt{10}+2;2-\sqrt{10}\}`