Toán Lớp 8: Tìm x, biet (x+1)-(x-5)(x+5)=22

Question

Toán Lớp 8:  Tìm x, biet (x+1)-(x-5)(x+5)=22

thuminhthong · Answer

Giải đáp:   $x=  dfrac{1+ sqrt{17}}{2}$ hoặc $x= dfrac{1- sqrt{17}}{2}$   Lời giải và giải thích chi tiết:    (x+1)-(x-5)(x+5) =22   &hArr;x+1 - (x^2 -5^2) =22   &hArr;x+1 -x^2 +5^2 =22   &hArr;-x^2 +x+1 +25-22=0   &hArr;-x^2 +x+4=0   &hArr;-x^2 + 2. 1/2 .x- 1/4 +17/4 =0   &hArr;17/4 -(x^2 -2. 1/2 .x + 1/4) =0   &hArr;17/4 - (x- 1/2)^2 =0   &hArr;(x- 1/2)^2 =17/4   &hArr;(x- 1/2)^2 =(&plusmn; ( sqrt{17})/2)^2   &hArr;x- 1/2 = &plusmn; ( sqrt{17})/2   &hArr; ( left[  begin{array}{l}x-  dfrac{1}{2} =  dfrac{ sqrt{17}}{2}  x-  dfrac{1}{2} = - dfrac{ sqrt{17}}{2} end{array}  right. )    &hArr; ( left[  begin{array}{l}x= dfrac{ sqrt{17}}{2}+ dfrac{1}{2}  x=- dfrac{ sqrt{17}}{2}+ dfrac{1}{2} end{array}  right. )    &hArr; ( left[  begin{array}{l}x= dfrac{1+ sqrt{17}}{2}  x= dfrac{1- sqrt{17}}{2} end{array}  right. )    Vậy $x=  dfrac{1+ sqrt{17}}{2}$ hoặc $x= dfrac{1- sqrt{17}}{2}$

tuanh · Answer

$ ( x + 1 ) - ( x - 5 ) ( x + 5 ) = 22$   &rArr; $ x+ 1 - x^2 + 25 = 22 $   &rArr; $ - x^2 + x + 4 = 0 $   &rArr;$ x^2 - x - 4 = 0 $   &rArr; $ x^2 - 2 . x .  dfrac{1}{2} +  dfrac{1}{4} -   dfrac{17}{4} = 0 $   &rArr; $ ( x -  dfrac{1}{2} )^2 -  dfrac{17}{4} = 0 $   &rArr; $ ( x -  dfrac{1}{2} )^2= dfrac{17}{4}$   &rArr; $ ( x -  dfrac{1}{2} )^2= ( &plusmn;  dfrac{ sqrt[]{17}}{2} )^2 $   &rArr;  ( left[  begin{array}{l}x- dfrac{1}{2}= dfrac{ sqrt[]{17}}{2}  x- dfrac{1}{2}= - dfrac{ sqrt[]{17}}{2} end{array}  right. )    &rArr;  ( left[  begin{array}{l}x= dfrac{1+ sqrt[]{17}}{2}  x= dfrac{1- sqrt[]{17}}{2} end{array}  right. )    Vậy $ x =  dfrac{1+ sqrt[]{17}}{2}$ hoặc $ x =  dfrac{1- sqrt[]{17}}{2}$