Tìm Min
A=\(\sqrt{x^2+x+1}+\sqrt{x^2-x-1}\)
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A= \(\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
=\(\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
=\(\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\)
\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|\)
=2.
dấu = khi và chỉ khi \(\left(\sqrt{x-1}+1\right).\left(1-\sqrt{x-1}\right)=0\)
\(A=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\)
\(\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\Leftrightarrow0\le x\le2\)
Vậy \(A_{min}=2\) tại \(0\le x\le2\)
ĐK: \(x\ge1\)
\(A=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\)
\(\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)
Đẳng thức xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\)
\(\Leftrightarrow1\le x\le2\)
a) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
b) \(\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\)
\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=\left|2\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow1\le x\le2\)
a) ĐKXĐ : \(3\le x\le7\)
Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)
\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)
Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)
a) \(\orbr{\orbr{\begin{cases}x\ge\sqrt{5}\\x\le-\sqrt{5}\end{cases}}}\) b)\(\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\)
c)\(\orbr{\begin{cases}\hept{\begin{cases}x\ge\sqrt{2}\\x\ne\sqrt{3}\end{cases}}\\\hept{\begin{cases}x\le-\sqrt{2}\\x\ne-\sqrt{3}\end{cases}}\end{cases}}\)
T có A=\(\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}\)
Áp dụng Bđt cô si
Taco\(A=\sqrt{x}+\dfrac{3}{\sqrt{x}}\ge2\sqrt{\dfrac{3}{\sqrt{x}}.\sqrt{x}}=2\sqrt{3}\)
Vậy\(Min_A=2\sqrt{3}\)
Dấu '=' xảy ra <=>x=0