Toán Lớp 8: (4x+1)^2-(2x+3)^2+5(x+2)^2+3(x-2)(x+2) = 500

Question

Toán Lớp 8:  (4x+1)^2-(2x+3)^2+5(x+2)^2+3(x-2)(x+2) = 500

chauhoangmy98 · Answer

Giải đáp:   x in{ frac{ sqrt{629}-2}{5}; frac{- sqrt{629}-2}{5}}   Lời giải và giải thích chi tiết:   (4x+1)&sup2;-(2x+3)&sup2;+5(x+2)&sup2;+3(x-2)(x+2)=500   &hArr;16x&sup2;+8x+1-(4x&sup2;+12x+9)+5(x&sup2;+4x+4)+3(x&sup2;-4)=500   &hArr;16x&sup2;+8x+1-4x&sup2;-12x-9+5x&sup2;+20x+20+3x&sup2;-12-500=0   &hArr;(16x&sup2;-4x&sup2;+5x&sup2;+3x&sup2;)+(8x-12x+20x)+(1-9+20-12-500)=0   &hArr;20x&sup2;+16x-500=0   &hArr;20(x&sup2;+4/5x-25)=0   &hArr;x&sup2;+4/5x-25=0   &hArr;x&sup2;+4/5x+4/25-629/25=0   &hArr;x&sup2;+4/5x+4/25=629/25   &hArr;x&sup2;+2.x. 2/5+(2/5)^2=629/25   &hArr;(x+2/5)^2=(( sqrt{629})/5)^2 &hArr;$ left[ begin{matrix} x+ dfrac{2}{5}= dfrac{ sqrt{629}}{5}   x+ dfrac{2}{5}=- dfrac{ sqrt{629}}{5} end{matrix} right.$ &hArr;$ left[ begin{matrix} x= dfrac{ sqrt{629}}{5}- dfrac{2}{5}   x=- dfrac{ sqrt{629}}{5}- dfrac{2}{5} end{matrix} right.$ &hArr;$ left[ begin{matrix} x= dfrac{ sqrt{629}-2}{5}   x= dfrac{- sqrt{629}-2}{5} end{matrix} right.$ Vậy x in{ frac{ sqrt{629}-2}{5}; frac{- sqrt{629}-2}{5}}