Toán CMR biểu thức sau luôn dương B = $2x^{2}$ + 7x + 10 12/07/2021 By Charlie CMR biểu thức sau luôn dương B = $2x^{2}$ + 7x + 10
$B=2x^2+7x+10$ $B=2(x^2+\frac{7}{2}x+5)$ $B=2(x+\frac{7}{4})^2+\frac{31}{8}$ Do $(x+\frac{7}{4})^2≥0⇒2(x+\frac{7}{4})^2+\frac{31}{8}≥\frac{31}{8}$ $⇒dpcm$ Trả lời
\(B=2x^2+7x+10\\=2(x^2+\dfrac{7}{2}x+5)\\=2(x^2+2.\dfrac{7}{4}x+\dfrac{49}{16}+\dfrac{31}{16})\\=2(x+\dfrac{7}{4})^2+\dfrac{31}{8}\) Vì \( (x+\dfrac{7}{4})^2\ge 0\\\to 2(x+\dfrac{7}{4})^2\ge 0\\\to B\ge \dfrac{31}{8}\\\to B>0\) Vậy biểu thức \(B=2x^2+7x+10\) luôn dương Trả lời
$B=2x^2+7x+10$
$B=2(x^2+\frac{7}{2}x+5)$
$B=2(x+\frac{7}{4})^2+\frac{31}{8}$
Do $(x+\frac{7}{4})^2≥0⇒2(x+\frac{7}{4})^2+\frac{31}{8}≥\frac{31}{8}$
$⇒dpcm$
\(B=2x^2+7x+10\\=2(x^2+\dfrac{7}{2}x+5)\\=2(x^2+2.\dfrac{7}{4}x+\dfrac{49}{16}+\dfrac{31}{16})\\=2(x+\dfrac{7}{4})^2+\dfrac{31}{8}\)
Vì \( (x+\dfrac{7}{4})^2\ge 0\\\to 2(x+\dfrac{7}{4})^2\ge 0\\\to B\ge \dfrac{31}{8}\\\to B>0\)
Vậy biểu thức \(B=2x^2+7x+10\) luôn dương