Toán Lớp 10: Giải phương trình: `2x^2+x+2=sqrt{3x+1}+sqrt{8x+1}`

Question

Toán Lớp 10:  Giải phương trình: 2x^2+x+2=sqrt{3x+1}+sqrt{8x+1}

ngoclanquynh · Answer

(2x^2 +x +2 = sqrt{3x+1} + sqrt{8x+1} qquad (x ge - dfrac{1}{8}) Leftrightarrow sqrt{3x +1}+ sqrt{8x+1} = 2x^2 +x+2 Leftrightarrow sqrt{3x+1} + sqrt{8x+1} -2x^2 -x -2x +2x -1 -1=0 Leftrightarrow ( sqrt{3x+1}-x-1)+( sqrt{8x+1} -2x-1)+ (2x-2x^2)=0 Leftrightarrow dfrac{(x-x^2)( sqrt{3x+1}-x-1)}{x-x^2}+ dfrac{(4x-4x^2)( sqrt{8x+1}-2x-1)}{4x-4x^2}+(2x-2x^2)=0 Leftrightarrow dfrac{(x-x^2)( sqrt{3x+1}-x-1)}{3x-2x-x^2 +1-1}+ dfrac{(4x-4x^2)( sqrt{8x+1}-2x-1)}{8x-4x-4x^2 +1-1}+(2x-2x^2)=0 Leftrightarrow dfrac{(x-x^2)( sqrt{3x+1}-x-1)}{3x+1-(x+1)^2}+ dfrac{(4x-4x^2)( sqrt{8x+1}-2x-1)}{8x+1-(2x+1)^2}+(2x-2x^2)=0 Leftrightarrow dfrac{(x-x^2)( sqrt{3x+1}-x-1)}{ sqrt{3x+1}^2 -(x+1)^2}+ dfrac{(4x+4x^2)( sqrt{8x+1}-2x-1)}{ sqrt{8x+1}^2-(2x+1)^2}+(2x-2x^2)=0 Leftrightarrow dfrac{(x-x^2)( sqrt{3x+1}-x-1)}{( sqrt{3x+1}+x+1)( sqrt{3x+1}-x-1)}+ dfrac{(4x-4x^2)( sqrt{8x+1}-2x-1)}{( sqrt{8x+1}+2x+1)( sqrt{8x+1}-2x-1)}+(2x-2x^2)=0 Leftrightarrow dfrac{x-x^2}{ sqrt{3x+1}+x+1}+ dfrac{4(x-x^2)}{ sqrt{8x+1}+2x+1}+2(x-x^2)=0 Leftrightarrow (x-x^2)( dfrac{1}{ sqrt{3x+1}+x+1}+ dfrac{4}{ sqrt{8x+1}+2x+1}+2)=0 ) Ta có ( dfrac{1}{ sqrt{3x+1}+x+1}+ dfrac{4}{ sqrt{8x+1}+2x+1}+2 ge 2 >0 ∀ x ) => x-x^2 =0 <=> x(1-x)=0 <=> x=0, quad x=1

hothaibinh · Answer

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