Toán Lớp 8: 2x^2-4x+2/x^2+2
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Toán Lớp 8: 2x^2-4x+2/x^2+2
Tìm Giá Trị lớn nhất
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Đặt $A=\dfrac{2x^2-4x+2}{x^2+2}$
$\to 3-A=3-\dfrac{2x^2-4x+2}{x^2+2}\\\to 3-A=\dfrac{3x^2+6 – 2x^2+4x-2}{x^2+2}\\\to 3-A=\dfrac{x^2 + 4x+4}{x^2+2}\\\to 3-A=\dfrac{(x+2)^2}{x^2+2}\ge 0∀x\\\to A\le 3∀x$
Dấu “$=$” xảy ra khi : $x+2=0↔x=-2$
Vậy $A_{max}=3↔x=-2$