huhu giúp em nhanh vs ạ, 10h là em phải có rồi, giúp em bài 2,3,4 ạ
View attachment 14522
$2)
\\A=\dfrac{2x^2+10}{x^2+11}=\dfrac{2(x^2+11)-12}{x^2+11}=2-\dfrac{12}{x^2+11}\geq 2-\dfrac{12}{11}=\dfrac{10}{11}$
Dấu '=' xảy ra khi $x=0$
$B=\dfrac{3x^2+2}{2x^2+3}=\dfrac{6x^2+4}{2(2x^2+3)}=\dfrac{3(2x^2+3)-5}{2(2x^2+3)}=\dfrac{3}{2}-\dfrac{5}{2(2x^2+3)}\geq \dfrac{3}{2}-\dfrac{5}{6}=\dfrac{2}{3}$
Dấu '=' xảy ra khi $x=0$
$C=x^2+5x+9=(x^2+5x+\dfrac{25}4)+\dfrac{11}4=(x+\dfrac{5}2)^2+\dfrac{11}4\geq \dfrac{11}4$
Dấu '=' xảy ra khi $x=\dfrac{-5}2$
$D=|x-2010|+|x+2017|=|2010-x|+|x+2017|\geq |2010-x+x+2017|=4027$
Dấu '=' xảy ra khi $(2010-x)(x+2017)\geq 0\Leftrightarrow -2017\leq x\leq 2010$
$E=|3x-5|+|3x-3|=|5-3x|+|3x-3|\geq |5-3x+3x-3|=2$
Dấu '=' xảy ra khi $(5-3x)(3x-3)\geq 0\Leftrightarrow 1\leq x\leq \dfrac{5}3$
$G=\dfrac{10\sqrt{x}+7}{5\sqrt{x}+14}=2-\dfrac{21}{5\sqrt{x}+14}\geq 2-\dfrac{21}{14}=\dfrac{1}{2}$
Dấu '=' xảy ra khi $x=0$
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