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Áp dụng BĐT Mincopxki:
\(P=\sqrt{\left(a^2\right)^2+1^2}+\sqrt{\left(b^2\right)^2+1^2}\ge\sqrt{\left(a^2+b^2\right)^2+\left(1+1\right)^2}\)
Ta xét:
\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)
\(a+b=\left(a+1\right)+\left(b+1\right)-2\ge2\sqrt{\left(a+1\right)\left(b+1\right)}-2=2.\frac{3}{2}-2=1\)
\(Đ\text{T}\Leftrightarrow a=b=\frac{1}{2}\)
\(\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=4\Leftrightarrow\sqrt{ab}+\sqrt{a}+\sqrt{b}=3\)
\(\text{Ta có:}M\ge a+b\Rightarrow2M+2\ge a+b+a+1+b+1\ge2\left(\sqrt{ab}+\sqrt{a}+\sqrt{b}\right)\left(\text{theo cô si}\right)=6\)
\(\Rightarrow M\ge2\left(\text{dấu "=" xảy ra khi:}a=b=1\right)\)
a) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)
Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)
\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)
Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
a)
P = \(\dfrac{1-\sqrt{a}+1+\sqrt{a}}{2\left(1-a\right)}-\dfrac{a^2+2}{\left(1-\sqrt{a}\right).\left(1+\sqrt{a}+a\right)}\)
P = \(\dfrac{2}{2\left(1-a\right)}-\dfrac{a^2+2}{\left(1-\sqrt{a}\right).\left(1+\sqrt{a}+a\right)}\)
P = \(\dfrac{1+\sqrt{a} +a-a^2-2}{\left(1-\sqrt{a}\right).\left(1+\sqrt{a}+a\right)}\)
P = \(\dfrac{a^2+a+\sqrt{a}-1}{\left(1-\sqrt{a}\right).\left(1+\sqrt{a}+a\right)}\)
P = \(\dfrac{\left(a^2+\sqrt{a}\right)+\left(a-1\right)}{\left(1-\sqrt{a}\right).\left(1+\sqrt{a}+a\right)}\)
P =
Hình như bạn viết nhầm đề, làm gì có số 9 ở đầu?
\(\frac{1}{1+a}+\frac{1}{1+b}\ge2\sqrt{\frac{1}{\left(1+a\right)\left(1+b\right)}}\)
\(\frac{a}{1+a}+\frac{b}{1+b}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\)
Cộng vế với vế: \(1\ge\frac{1+\sqrt{ab}}{\sqrt{\left(1+a\right)\left(1+b\right)}}\Leftrightarrow\left(1+a\right)\left(1+b\right)\ge\left(1+\sqrt{ab}\right)^2\)
Áp dụng xuống dưới ta có:
\(M\ge\left(1+\sqrt{b}\right)^2\left(1+\frac{4}{\sqrt{b}}\right)^2=\left(5+\frac{4}{\sqrt{b}}+\sqrt{b}\right)^2\ge\left(5+2\sqrt{\frac{4\sqrt{b}}{\sqrt{b}}}\right)^2=81\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=4\\a=2\end{matrix}\right.\)
dùng minscopxki bạn