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a) \(A=\sqrt{4x^2+4x+2}=\sqrt{4x^2+4x+1+1}=\sqrt{\left(2x+1\right)^2+1}\)
Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)
\(\Rightarrow A\ge\sqrt{1}=1\)
Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)
b) \(B=\sqrt{2x^2-4x+5+1}=\sqrt{2x^2-4x+2+3+1}=\sqrt{2\left(x^2-2x+1\right)+4}\)
\(=\sqrt{2\left(x-1\right)^2+4}\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow B\ge\sqrt{4}=2\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(minB=2\Leftrightarrow x=1\)
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ta có :
\(\sqrt{x^2+2x+1}+\sqrt{x^2+4x+4}=\left|x+1\right|+\left|x+2\right|\ge\left|x+1-x-2\right|=1\)
Dấu bằng xảy ra khi : \(\left(x+1\right)\left(x+2\right)\le0\Leftrightarrow-2\le x\le-1\)
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a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)
\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2
\(B=\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}=\sqrt{2\left[x^2-2x+1+\dfrac{3}{2}\right]}\)
\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1
\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1
\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2
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1.
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$A=|x+2|+|x+3|=|x+2|+|-x-3|\geq |x+2-x-3|=1$
Vậy GTNN của $A$ là $1$. Giá trị này đạt tại $(x+2)(-x-3)\geq 0$
$\Leftrightarrow (x+2)(x+3)\leq 0$
$\Leftrightarrow -3\leq x\leq -2$
2. ĐKXĐ: $x\geq 1$
\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}\)
\(=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|\geq |\sqrt{x-1}+1+1-\sqrt{x-1}|=2\)
Vậy gtnn của $B$ là $2$. Giá trị này đạt tại $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$
$\Leftrightarrow 1-\sqrt{x-1}\geq 0$
$\Leftrightarrow 0\leq x\leq 2$
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Lời giải:
Đặt $\sqrt{y}=b(b\geq 0)\Rightarrow y=b^2$
$M=2x^2+5b^2-4xb-4x-8b+2036$
$=2(x^2+b^2-2xb)+3b^2-4x-8b+2036$
$=2(x-b)^2-4(x-b)+3b^2-12b+2036$
$=2(x-b)^2-4(x-b)+2+3(b^2-4b+4)+2022$
$=2[(x-b)^2-2(x-b)+1]+3(b-2)^2+2022$
$=2(x-b-1)^2+3(b-2)^2+2022\geq 2022$
Vậy $M_{\min}=2022$
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a) Ta có: \(F=\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge\sqrt{1}=1\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy Min(F) = 1 khi x=2
b) \(D=\sqrt{2x^2-4x+10}=\sqrt{2\left(x-1\right)^2+8}\ge\sqrt{8}=2\sqrt{2}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(Min\left(D\right)=2\sqrt{2}\Leftrightarrow x=1\)
c) \(G=\sqrt{2x^2-6x+5}=\sqrt{2\left(x-\frac{3}{2}\right)^2+\frac{1}{2}}\ge\sqrt{\frac{1}{2}}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy \(Min\left(G\right)=\frac{\sqrt{2}}{2}\Leftrightarrow x=\frac{3}{2}\)