tìm x:::::::: (x-2).(x+7)=17

Đáp án:

`x=(\sqrt{149}-5)/2` hoặc `x=(-\sqrt{149}-5)/2.`

Giải thích các bước giải:

`(x-2).(x+7)=17`

`<=>x.(x+7)-2.(x+7)=17`

`<=>x.x+x.7-2.x-2.7=17`

`<=>x^1 . x^1+7.x-2.x-14=17`

`<=>x^(1+1)+7.x-2.x-14=17`

`<=>x^2+7.x-2.x-14=17`

`<=>x^2+7.x-2.x-14-17=0`

`<=>x^2+(7.x-2.x)+[(-14)-17]=0`

`<=>x^2+5x-31=0`

`<=>x^2+2 . 5/2x-31+(5/2)^2-(5/2)^2=0`

`<=>[x^2+2 . 5/2x+(5/2)^2]-31-(5/2)^2=0`

`<=>(x+5/2)^2-31-(5/2)^2=0`

`<=>(x+5/2)^2=31+(5/2)^2`

`<=>(x+5/2)^2=31+5^2/2^2`

`<=>(x+5/2)^2=31+25/4`

`<=>(x+5/2)^2=31/1+25/4`

`<=>(x+5/2)^2=124/4+25/4`

`<=>(x+5/2)^2=(124+25)/4`

`<=>(x+5/2)^2=149/4`

`<=>x+5/2=+-\sqrt{149/4}`

`<=>x=+-\sqrt{149/4}-5/2`

`<=>x=+-(\sqrt{149})/2-5/2`

`<=>x=(+-\sqrt{149}-5)/2`

`<=>x=(\sqrt{149}-5)/2` hoặc `x=(-\sqrt{149}-5)/2.`

Đáp án:

Giải thích các bước giải:

`\color{red}{\text{@sharinganvanhoadong}}`

 `(x-2)*(x+7) = 17`

 `x^2+7x-2x-14 = 17`

 `x^2+5x-31           = 0`

`x*(x+2,5)+2,5x + 6,25 - 6,25 - 31 = 0`

 `x.(x+2,5)+2,5.(x+2,5) - 37,25 = 0`

 `x*(x+2,5)+2,5x+2,5^2-37,25 = 0`

 `(x+2,5)^2 = 37,25`

 `(x+2,5)^2 = 149/4`

`=> x+2,5 = +- \sqrt{149}/2`

`=> x        = +- (\sqrt{149})/2 - 2,5`

`=> x        = +- (\sqrt{149})/2 - 5/2`

`=> x        = (+- \sqrt{149}-5)/2`
Vậy x in {( \sqrt{149}-5)/2; (- \sqrt{149}-5)/2`