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\(P=\sqrt{1+\left(a^2\right)^2}+\sqrt{1+\left(b^2\right)^2}\ge\sqrt{\left(1+1\right)^2+\left(a^2+b^2\right)^2}\)
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
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Mới làm xong :) Câu hỏi của khánh khang zen - Toán lớp 10 | Học trực tuyến
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/ ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
\(A=\left[\frac{1}{\sqrt{x}-1}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}+4}{\sqrt{x}-1}\right]\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+1}\)
b/
Ta có: \(A=\frac{1}{\sqrt{x}+1}\ge1\)
Vậy Min A = 1 .Dấu "=" xảy ra khi x = 0
a , rút gọn : A= \(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{x-1}\right):\left(2-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)
A= \(\left(\frac{1\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)
A= \(\left(\frac{\sqrt{x}+1+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+4}{\sqrt{x}-1}\right)\)
A= \(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
A=\(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
A = \(\frac{1}{\sqrt{x}+1}\)
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 \(\left\{{}\begin{matrix}b=4\\a=2\end{matrix}\right.\)