a)5x(x-3)-x-3=0 b)x^2+3x+x-3=0 c)8x(x+5)-x-5=0 giúp mik nhanh vs ak mik cảm ơn 13/07/2021 Bởi Serenity a)5x(x-3)-x-3=0 b)x^2+3x+x-3=0 c)8x(x+5)-x-5=0 giúp mik nhanh vs ak mik cảm ơn
`a)` `5x(x-3)-x-3=0` `<=> 5x^2-15x-x-3=0` `<=> 5x^2-16x-3=0` `<=> x^2-16/5x-3/5=0` `<=> x^2-2 . x . 8/5+(8/5)^2-3/5-(8/5)^2=0` `<=> (x-8/5)^2-79/25=0` `<=> (x-8/5)^2=79/25` `<=>` \(\left[ \begin{array}{l}x-\dfrac{8}{5}=\dfrac{\sqrt{79}}{5}\\x-\dfrac{8}{5}=-\dfrac{\sqrt{79}}{5}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{8+\sqrt{79}}{5}\\x=\dfrac{8-\sqrt{79}}{5}\end{array} \right.\) Vậy `S={(8+-sqrt(79))/5}` `b)` `x^2+3x+x-3=0` `<=> x^2+4x-3=0` `<=> x^2+4x+4-7=0` `<=> (x+2)^2=7` `<=>` \(\left[ \begin{array}{l}x+2=\sqrt{7}\\x+2=-\sqrt{7}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2+\sqrt{7}\\x=-2-\sqrt{7}\end{array} \right.\) Vậy `S={-2+-sqrt(7)}` `c)` `8x(x+5)-x-5=0` `<=> 8x(x+5)-(x+5)=0` `<=> (8x-1)(x+5)=0` `<=>` \(\left[ \begin{array}{l}8x-1=0\\x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{8}\\x=-5\end{array} \right.\) Vậy `S={1/8;-5}` Bình luận
Đáp án: a)Vậy S={3,$\frac{1}{5}$} b)Vậy S={-3,1} c)Vậy S={-5,$\frac{1}{8}$} Giải thích các bước giải: a)5x(x-3)-x+3=0 ⇔5x(x-3)-(x-3)=0 ⇔(x-3)(5x-1)=0 ⇒x-3=0 hoặc 5x-1=0 x=3 hoặc x=$\frac{1}{5}$ Vậy S={3,$\frac{1}{5}$} b)$x^{2}$+3x-x-3=0 ⇔$x^{2}$+3x-(x+3)=0 ⇔x(x+3)-(x+3)=0 ⇔(x+3)(x-1)=0 ⇒x+3=0 hoặc x-1=0 x=-3 hoặc x=1 Vậy S={-3,1} c)8x(x+5)-x-5=0 ⇔c)8x(x+5)-(x+5)=0 ⇔(x+5)(8x-1)=0 ⇒x+5=0 hoặc 8x-1=0 x=-5 hoặc x=$\frac{1}{8}$ Vậy S={-5,$\frac{1}{8}$} Bình luận
`a)`
`5x(x-3)-x-3=0`
`<=> 5x^2-15x-x-3=0`
`<=> 5x^2-16x-3=0`
`<=> x^2-16/5x-3/5=0`
`<=> x^2-2 . x . 8/5+(8/5)^2-3/5-(8/5)^2=0`
`<=> (x-8/5)^2-79/25=0`
`<=> (x-8/5)^2=79/25`
`<=>` \(\left[ \begin{array}{l}x-\dfrac{8}{5}=\dfrac{\sqrt{79}}{5}\\x-\dfrac{8}{5}=-\dfrac{\sqrt{79}}{5}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{8+\sqrt{79}}{5}\\x=\dfrac{8-\sqrt{79}}{5}\end{array} \right.\)
Vậy `S={(8+-sqrt(79))/5}`
`b)`
`x^2+3x+x-3=0`
`<=> x^2+4x-3=0`
`<=> x^2+4x+4-7=0`
`<=> (x+2)^2=7`
`<=>` \(\left[ \begin{array}{l}x+2=\sqrt{7}\\x+2=-\sqrt{7}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2+\sqrt{7}\\x=-2-\sqrt{7}\end{array} \right.\)
Vậy `S={-2+-sqrt(7)}`
`c)`
`8x(x+5)-x-5=0`
`<=> 8x(x+5)-(x+5)=0`
`<=> (8x-1)(x+5)=0`
`<=>` \(\left[ \begin{array}{l}8x-1=0\\x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{8}\\x=-5\end{array} \right.\)
Vậy `S={1/8;-5}`
Đáp án:
a)Vậy S={3,$\frac{1}{5}$}
b)Vậy S={-3,1}
c)Vậy S={-5,$\frac{1}{8}$}
Giải thích các bước giải:
a)5x(x-3)-x+3=0
⇔5x(x-3)-(x-3)=0
⇔(x-3)(5x-1)=0
⇒x-3=0 hoặc 5x-1=0
x=3 hoặc x=$\frac{1}{5}$
Vậy S={3,$\frac{1}{5}$}
b)$x^{2}$+3x-x-3=0
⇔$x^{2}$+3x-(x+3)=0
⇔x(x+3)-(x+3)=0
⇔(x+3)(x-1)=0
⇒x+3=0 hoặc x-1=0
x=-3 hoặc x=1
Vậy S={-3,1}
c)8x(x+5)-x-5=0
⇔c)8x(x+5)-(x+5)=0
⇔(x+5)(8x-1)=0
⇒x+5=0 hoặc 8x-1=0
x=-5 hoặc x=$\frac{1}{8}$
Vậy S={-5,$\frac{1}{8}$}