Tìm GTNN
\(H=\left|3000-x\right|+\left|2025-x\right|\)
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c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
#)Giải :
\(H=\left|x-4\right|\left(2-\left|x-4\right|\right)\)
\(=-\left(\left|x-4\right|\right)^2+2\left|x-4\right|\)
\(=-\left[\left(\left|x-4\right|\right)^2-2\left|x-4\right|\right]\)
\(=-\left[\left(\left|x-4\right|-1\right)^2-1\right]\le0\)
Dấu ''='' xảy ra khi \(\left|x-4\right|-1=0\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
Bạn ơi mình nghĩ GTNN phải là -1
Vì ko có GTNN bằng 0
\(H\ge\left|\left(x+2\right)+\left(4-x\right)\right|\)
\(\Rightarrow H\ge2\)
\(\Rightarrow Hmin=2\Leftrightarrow\left|x-2\right|+\left|x-4\right|=2\)
NẾU \(x< 2\):
\(\left|2-x\right|+\left|4-x\right|=2\)
\(\Leftrightarrow2-x+4-x=2\)
\(\Leftrightarrow6-2x=2\Leftrightarrow x=2\left(KTM\right)\)
NẾU :\(2\le x\le4\)
\(\left|x-2\right|+\left|4-x\right|=2\)
\(\Leftrightarrow x-2+4-x=2\left(TM\right)\)
NẾU :\(x>4\)
\(\Leftrightarrow\left(x-2\right)+\left(x-4\right)=2\)
\(\Leftrightarrow2x-6=2\Rightarrow x=4\left(KTM\right)\)
VẬY:\(Hmin=2\)khi\(2\le x\le4\)
A=(x^2+5x-6)(x^2+5x+6)
=(x^2+5x)^2-36>=-36
Dấu = xảy ra khi x=0 hoặc x=-5
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):
\(A=\left|x-3\right|+\left|x-1\right|+\left|x+1\right|+\left|x+3\right|\)
\(=\left|3-x\right|+\left|x+3\right|+\left|1-x\right|+\left|x+1\right|\)
\(\ge\left|3-x+x+3\right|+\left|1-x+x+1\right|=8\)
\(minA=8\Leftrightarrow\left\{{}\begin{matrix}\left(3-x\right)\left(x+3\right)\ge0\\\left(1-x\right)\left(x+1\right)\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le1\)
Ta có tính chất :
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\rightarrow A=\left|x+5\right|+\left|x+2\right|+\left|x-7\right|+\left|x-8\right|\ge\left|x+5+x+2+x-7+x-8\right|\)
\(\rightarrow A\ge\left|4x-8\right|\)
Vì \(\left|4x-8\right|\ge0\forall x\in R\) nên :
\(\rightarrow A\ge0\forall x\in R\)
Dấu "= " xảy ra khi :
\(\left|4x-8\right|=0\) \(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
Vậy \(A_{min}=0\Leftrightarrow x=2\)