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A=\(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+3\right)^2}\)=|x-1|+|x+3|=|1-x|+|x+3|
Áp dụng bđt |a|+|b|\(\ge\)|a+b| ta được: A=|1-x|+|x+3|\(\ge\)|1-x+x+3|=4
Dấu "=" xảy ra khi (1-x)(x+3)\(\ge\)0 <=> \(-3\le x\le1\)
Vậy Amin=4 khi \(-3\le x\le1\)
A = \(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}\)
= \(\sqrt{\left(1-x\right)^2}+\sqrt{\left(x+3\right)^2}\)
= 1 - x + x + 3
= 4
\(C=\sqrt{2x^2-6x+5}=\sqrt{2\left(x^2-3x+\frac{9}{4}\right)+\frac{1}{2}}\)
\(C=\sqrt{2\left(x^2-2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2\right)+\frac{1}{2}}=\sqrt{2\left(x-\frac{3}{2}\right)^2+\frac{1}{2}}\ge\frac{1}{2}\)
Vậy GTNN của C là \(\frac{1}{2}\) \(\Leftrightarrow x=\frac{3}{2}\)
mình nhầm. thay GTNN \(\frac{1}{2}\)thành \(\sqrt{\frac{1}{2}}\)
a) \(\sqrt{x}-x=-\left(x-\sqrt{x}\right)\)
\(=-\left[\left(\sqrt{x}\right)^2-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}\right]+\frac{1}{4}\)
\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy GTLN của bt là \(\frac{1}{4}\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)
\(\sqrt{2}A=\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\)
\(\sqrt{2}A=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\)
Áp dụng BĐT \(\sqrt{A^2+B^2}+\sqrt{C^2+D^2}\ge\sqrt{\left(A+C\right)^2+\left(B+D\right)^2}\)
=>\(\sqrt{2}A\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\)
=>\(A\ge\sqrt{13}\)
Dấu bằng xảy ra<=> \(\frac{2x-1}{3}=\frac{2x-2}{2}\)
<=>.........
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 \(minB=2\Leftrightarrow x=1\)
\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
\(=\sqrt{\left(1-3x\right)^2}+\sqrt{\left(3x-2\right)^2}\)
\(=\left|1-3x\right|+\left|3x-2\right|\)
\(\ge\left|1-3x+3x-2\right|=\left|-1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(1-3x\right)\left(3x-2\right)\ge0\Leftrightarrow\frac{1}{3}\le x\le\frac{2}{3}\)
Vậy \(A_{min}=1\) tại \(\frac{1}{3}\le x\le\frac{2}{3}\)
Ta có \(A=\sqrt{2x^2+6x+5}\)
\(=\sqrt{2\left(x^2+3x+\dfrac{5}{2}\right)}\)
\(=\sqrt{2\left(x^2+3x+\dfrac{9}{4}+\dfrac{1}{4}\right)}\)
\(=\sqrt{2\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}\right]}\ge\sqrt{2.\dfrac{1}{4}}=\dfrac{\sqrt{2}}{2}\)
Vậy GTNN của A là \(\dfrac{\sqrt{2}}{2}\) khi \(\left(x+\dfrac{3}{2}\right)^2=0\Leftrightarrow x=-\dfrac{3}{2}\)
Học tốt nhé