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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:
\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)
\(a+b=\left(a+1\right)+\left(b+1\right)-2\ge2\sqrt{\left(a+1\right)\left(b+1\right)}-2=2.\frac{3}{2}-2=1\)
\(Đ\text{T}\Leftrightarrow a=b=\frac{1}{2}\)
Mới làm xong :) Câu hỏi của khánh khang zen - Toán lớp 10 | Học trực tuyến
\(\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=4\Leftrightarrow\sqrt{ab}+\sqrt{a}+\sqrt{b}=3\)
\(\text{Ta có:}M\ge a+b\Rightarrow2M+2\ge a+b+a+1+b+1\ge2\left(\sqrt{ab}+\sqrt{a}+\sqrt{b}\right)\left(\text{theo cô si}\right)=6\)
\(\Rightarrow M\ge2\left(\text{dấu "=" xảy ra khi:}a=b=1\right)\)
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 \(\left\{{}\begin{matrix}b=4\\a=2\end{matrix}\right.\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
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 \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(P=\sqrt{1+\left(a^2\right)^2}+\sqrt{1+\left(b^2\right)^2}\ge\sqrt{\left(1+1\right)^2+\left(a^2+b^2\right)^2}\)
Mà \(a+b=\left(a+1\right)+\left(b+1\right)-2\ge2\sqrt{\left(a+1\right)\left(b+1\right)}-2=2.\sqrt{\frac{9}{4}}-2=1\)
Và \(a^2+b^2\ge\frac{\left(a+b\right)^2}{4}\ge\frac{1}{4}\)
=>\(P\ge\sqrt{4+\frac{1}{4}}=\sqrt{\frac{17}{4}}\)
Pmin =\(\sqrt{\frac{17}{4}}\) khi a=b =1/2