Đáp án+giải thích các bước giải:
$A=2x^2+2x+17$
$=2(x^2+x+\dfrac{17}{2}$
$=2[(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4})+\dfrac{33}{4}]$
$=2(x+\dfrac{1}{2})^2+\dfrac{33}{2}$
do $(x+\dfrac{1}{2})^2≥0 ∀x$
⇒ $2(x+\dfrac{1}{2})^2+\dfrac{33}{2}≥\dfrac{33}{2} ∀x$
Dấu"=" xảy ra khi
$x+\dfrac{1}{2}=0$
$⇒x=\dfrac{-1}{2}$
$B=3x^2+5x+1$
$=3(x^2+\dfrac{5}{3}x+\dfrac{1}{3})$
$=3[(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36})-\dfrac{13}{36}]$
$=3(x+\dfrac{5}{6})-\dfrac{13}{12}$
do $(x+\dfrac{5}{6})^2≥0∀x$
⇒ $3(x+\dfrac{5}{6})^2-\dfrac{13}{12}≥-\dfrac{13}{12}∀x$
Dấu"=" xảy ra khi
$x+\dfrac{5}{6}=0$
$⇒x=\dfrac{-5}{6}$
$C=7x^2-3x+11$
$=7(x^2-\dfrac{3}{7}x+\dfrac{11}{7})$
$=7[(x^2-2.x.\dfrac{3}{14}+\dfrac{9}{196})+\dfrac{299}{196}]$
$=7(x-\dfrac{3}{14})^2+\dfrac{299}{28}$
do $(x-\dfrac{3}{14})^2≥0∀x$
⇒ $7(x-\dfrac{3}{14})^2+\dfrac{299}{28}≥\dfrac{299}{28}∀x$
Dấu "=" xảy ra khi
$x-\dfrac{3}{14}=0$
$⇒x=\dfrac{3}{14}$
$D=5x^2-7x+13$
$=5(x^2-\dfrac{7}{5}x+\dfrac{13}{5}$
$=5[(x^2-2.x.\dfrac{7}{10}+\dfrac{49}{100})+\dfrac{211}{100}]$
$=5(x-\dfrac{7}{10})^2+\dfrac{211}{20}$
do $(x-\dfrac{7}{10})^2≥0∀x$
⇒ $5(x-\dfrac{7}{10})^2 + \dfrac{211}{20}≥\dfrac{211}{20}x$
Dấu "=" xảy ra khi
$x-\dfrac{7}{10}=0$
$⇒x=\dfrac{7}{10}$
