Đáp án:
Giải thích các bước giải:
$F=|2x-\dfrac{3}{5}|+1,(3)$
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Ta có: $|2x-\dfrac{3}{5}|≥0,∀x$
$⇒F=|2x-\dfrac{3}{5}|+1,(3)≥1,(3)$
$⇒Fmin=1,(3)$ khi $2x-\dfrac{3}{5}=0⇔x=\dfrac{3}{10}$
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$H=(x-\dfrac{2}{7})^{2006}+(0,2-\dfrac{1}{5}y)^{2004}+(-2)^{2005}$
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$=(x-\dfrac{2}{7})^{2006}+(0,2-\dfrac{1}{5}y)^{2004}-2^{2005}$
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Ta có: $(x-\dfrac{2}{7})^{2006}≥0,∀x; (0,2-\dfrac{1}{5}y)^{2004}≥0,∀y$
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$⇒H=(x-\dfrac{2}{7})^{2006}+(0,2-\dfrac{1}{5}y)^{2004}-2^{2005}≥-2^{2005}$
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$⇒Hmin=-2^{2005}$ khi $\left \{ {{x-\frac{2}{7}=0} \atop {0,2-\frac{1}{5}y=0}} \right.$⇔ $\left \{ {{x=\frac{2}{7}} \atop {y=1}} \right.$
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$G=|x-3|+|x+\dfrac{3}{2}|=|x-3|+|-x-\dfrac{3}{2}|≥|x-3-x-\dfrac{3}{2}|=\dfrac{9}{2}$
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$⇒Gmin=\dfrac{9}{2}$ khi $(x-3)(-x-\dfrac{3}{2})≥0$
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$TH1:$$\left \{ {{x-3≥0} \atop {-x-\frac{3}{2}≥0}} \right.$ ⇔$\left \{ {{x≥3} \atop {\frac{-3}{2}≥x}} \right.$ ⇔ không tồn tại $x$
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$TH2:$$\left \{ {{x-3≤0} \atop {-x-\frac{3}{2}≤0}} \right.$ ⇔$\left \{ {{x≤3} \atop {\frac{-3}{2}≤x}} \right.$ ⇔ $-\frac{3}{2}≤x≤3$
