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F=\(\sqrt{x^2+2019}\)
=>\(F^2=x^2+2019 =>x^2+2019\)≥2019
=> \(F^2 \)min=2019=>F min=\(\sqrt{2019}\)<=>x=0
G=\(\sqrt{x^2-x+1}\)=\(\sqrt{x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}}\)=\(\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\) \(\ge\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)
Dấu "=" xảy ra <=> x=\(\frac{1}{2}\)
Vậy minG=\(\frac{\sqrt{3}}{2}\) <=> x\(=\frac{1}{2}\)
Câu 1:
Áp dụng BĐT Cô-si:
\(A=\sqrt{\left(2-x\right)\left(2+x\right)}\le\frac{2-x+2+x}{2}=2\)
Dấu "=" xảy ra \(\Leftrightarrow2-x=2+x\Leftrightarrow x=0\)
Câu 2:
\(B=\sqrt{-x^2+x+\frac{1}{4}}\)
\(B=\sqrt{-\left(x^2-x-\frac{1}{4}\right)}\)
\(B=\sqrt{-\left(x^2-x+\frac{1}{4}-\frac{1}{2}\right)}\)
\(B=\sqrt{-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\right]}\)
\(B=\sqrt{\frac{1}{2}-\left(x-\frac{1}{2}\right)^2}\le\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
a) \(\sqrt{x-2}+\sqrt{16x-32}=10\)
\(\Rightarrow\sqrt{x-2}+4\sqrt{x-2}=10\)
\(\Rightarrow5\sqrt{x-2}=10\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
b) \(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)
ĐK \(x\ge\dfrac{1}{2}\)
\(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)
\(\left(\sqrt{x+\sqrt{2x-1}}\right)^2=\left(5\sqrt{2}\right)^2\)
\(\left|x+\sqrt{2x-1}\right|=50\)
\(\sqrt{2x-1}=50-x\)
\(\left(\sqrt{2x-1}\right)^2=\left(50-x\right)^2\)
\(\left|2x-1\right|=x^2-100x+2500\)
\(2x-1=x^2-100x+2500\)
\(x=41\)
a) ĐK: \(x\geq 0\)
Ta có: \(\sqrt{x}+\sqrt{x+1}=1\Leftrightarrow \sqrt{x}+\sqrt{x+1}-1=0\)
\(\Leftrightarrow \sqrt{x}+\frac{(x+1)-1}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow \sqrt{x}+\frac{x}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow \sqrt{x}\left(1+\frac{\sqrt{x}}{\sqrt{x+1}+1}\right)=0\)
Thấy rằng \(1+\frac{\sqrt{x}}{\sqrt{x+1}+1}>0, \forall x\geq 0\Rightarrow 1+\frac{\sqrt{x}}{\sqrt{x+1}+1}\neq 0\)
Do đó \(\sqrt{x}=0\Rightarrow x=0\) (thỏa mãn)
b) ĐK: \(x\geq 1\)
Ta thấy với mọi \(x\geq 1\) thì:\(\left\{\begin{matrix} \sqrt{x+4}\geq \sqrt{1+4}>2 \\ \sqrt{x-1}\geq 0\end{matrix}\right.\)
\(\Rightarrow \sqrt{x+4}+\sqrt{x-1}>2\)
Do đó pt \(\sqrt{x+4}+\sqrt{x-1}=2\) vô nghiệm