Toán Lớp 8: Tìm `x`để: `x^2 - x + 1/4 = 2x^3(1-2x)^2` -

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Toán Lớp 8: Tìm `x`để: `x^2 – x + 1/4 = 2x^3(1-2x)^2`

Cát Linh Tháng Sáu 21, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Tìm xđể:
x^2 – x + 1/4 = 2x^3(1-2x)^2

Comments ( 1 )

truongthanhhung
21 Tháng Sáu, 2022 at 9:17 sáng

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\(x^2 -x +\dfrac{1}{4}=2x^3(1-2x)^2\\\Leftrightarrow 4(x^2 -x+\dfrac{1}{4})=4[2x^3(1-2x)^2]\\\Leftrightarrow 4x^2 -4x+1=4(2x^3-8x^4+8x^5)\\\Leftrightarrow 4x^2 -4x+1-(8x^3 -32x^4+32x^5)=0\\\Leftrightarrow 4x(x-1)+1-8x^3+32x^4 (1-x)=0\\\Leftrightarrow 4x(x-1)+1-8x^3 -32x^4(x-1)=0\\\Leftrightarrow 4x(x-1)(1-8x^3)+1-8x^3=0\\\Leftrightarrow (1-8x^3)[4x(x-1)+1]=0\\\Leftrightarrow(1-8x^3)(2x-1)^2=0\\\Leftrightarrow \left[ \begin{array}{l}1-8x^3=0\\(2x-1)^2=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x^3=\dfrac{1}{8}\\2x=1\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{1}{2}\end{array} \right.\)

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