Toán tìm x biết: a) x^2 – 3x – 1 = 0 b)x^2 + 5x – 1 = 0 07/07/2021 By Ruby tìm x biết: a) x^2 – 3x – 1 = 0 b)x^2 + 5x – 1 = 0
Đáp án: `a)x^2-3x-1=0` `<=>x^2-3x=1` `<=>x^2-2*x*3/2+9/4=1+9/4` `<=>(x-3/2)^2=13/4` `<=>` \(\left[ \begin{array}{l}x-\dfrac32=\dfrac{\sqrt{13}}{2}\\x-\dfrac32=-\dfrac{\sqrt{13}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{13}+3}{2}\\x=\dfrac{-\sqrt{13}+3}{2}\end{array} \right.\) Vậy phương trình có tập nghiệm `S={(-\sqrt{13}+3)/2,(\sqrt{13}+3)/2}`. `b)x^2+5x-1=0` `<=>x^2+5x=1` `<=>x^2+2*x*5/2+25/4=1+25/4` `<=>(x+5/2)^2=29/4` `<=>` \(\left[ \begin{array}{l}x+\dfrac52=\dfrac{\sqrt{29}}{2}\\x+\dfrac52=-\dfrac{\sqrt{29}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{29}-5}{2}\\x=\dfrac{-\sqrt{29}-5}{2}\end{array} \right.\) Vậy phương trình có tập nghiệm `S={(-\sqrt{29}-5)/2,(\sqrt{29}-5)/2}`. Trả lời
Đáp án: `a) \ \ S={(3+sqrt(13))/2;(3-sqrt(13))/2}` `b) \ \ S={(sqrt(29)-5)/2;(-sqrt(29)-5)/2}` Giải thích các bước giải: `a) \ \ x^2-3x-1=0` `<=> x^2 – 2 . x . 3/2 + (3/2)^2 – 1 – (3/2)^2=0` `<=> (x-3/2)^2-13/4=0` `<=> (x-3/2)^2=13/4` `<=>` \(\left[ \begin{array}{l}x-\dfrac{3}{2}=\dfrac{\sqrt{13}}{2}\\x-\dfrac{3}{2}=-\dfrac{\sqrt{13}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{13}+3}{2}\\x=\dfrac{-\sqrt{13}+3}{2}\end{array} \right.\) Vậy `S = {(3+sqrt(13))/2;(3-sqrt(13))/2}` `b) \ \ x^2+5x-1=0` `<=> x^2+2.x . 5/2 +(5/2)^2-1-(5/2)^2=0` `<=> (x+5/2)^2-29/4=0` `<=> (x+5/2)^2=29/4` `<=>` \(\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt{29}}{2}\\x+\dfrac{5}{2}=-\dfrac{\sqrt{29}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{29}-5}{2}\\x=\dfrac{-\sqrt{29}-5}{2}\end{array} \right.\) Vậy `S={(sqrt(29)-5)/2;(-sqrt(29)-5)/2}` Trả lời
Đáp án:
`a)x^2-3x-1=0`
`<=>x^2-3x=1`
`<=>x^2-2*x*3/2+9/4=1+9/4`
`<=>(x-3/2)^2=13/4`
`<=>` \(\left[ \begin{array}{l}x-\dfrac32=\dfrac{\sqrt{13}}{2}\\x-\dfrac32=-\dfrac{\sqrt{13}}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{13}+3}{2}\\x=\dfrac{-\sqrt{13}+3}{2}\end{array} \right.\)
Vậy phương trình có tập nghiệm `S={(-\sqrt{13}+3)/2,(\sqrt{13}+3)/2}`.
`b)x^2+5x-1=0`
`<=>x^2+5x=1`
`<=>x^2+2*x*5/2+25/4=1+25/4`
`<=>(x+5/2)^2=29/4`
`<=>` \(\left[ \begin{array}{l}x+\dfrac52=\dfrac{\sqrt{29}}{2}\\x+\dfrac52=-\dfrac{\sqrt{29}}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{29}-5}{2}\\x=\dfrac{-\sqrt{29}-5}{2}\end{array} \right.\)
Vậy phương trình có tập nghiệm `S={(-\sqrt{29}-5)/2,(\sqrt{29}-5)/2}`.
Đáp án:
`a) \ \ S={(3+sqrt(13))/2;(3-sqrt(13))/2}`
`b) \ \ S={(sqrt(29)-5)/2;(-sqrt(29)-5)/2}`
Giải thích các bước giải:
`a) \ \ x^2-3x-1=0`
`<=> x^2 – 2 . x . 3/2 + (3/2)^2 – 1 – (3/2)^2=0`
`<=> (x-3/2)^2-13/4=0`
`<=> (x-3/2)^2=13/4`
`<=>` \(\left[ \begin{array}{l}x-\dfrac{3}{2}=\dfrac{\sqrt{13}}{2}\\x-\dfrac{3}{2}=-\dfrac{\sqrt{13}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{13}+3}{2}\\x=\dfrac{-\sqrt{13}+3}{2}\end{array} \right.\)
Vậy `S = {(3+sqrt(13))/2;(3-sqrt(13))/2}`
`b) \ \ x^2+5x-1=0`
`<=> x^2+2.x . 5/2 +(5/2)^2-1-(5/2)^2=0`
`<=> (x+5/2)^2-29/4=0`
`<=> (x+5/2)^2=29/4`
`<=>` \(\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt{29}}{2}\\x+\dfrac{5}{2}=-\dfrac{\sqrt{29}}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{29}-5}{2}\\x=\dfrac{-\sqrt{29}-5}{2}\end{array} \right.\)
Vậy `S={(sqrt(29)-5)/2;(-sqrt(29)-5)/2}`