cho (1+a)(1+b)=\(\dfrac{9}{4}\). tìm min P=\(\sqrt{1+a^4}+\sqrt{1+b^4}\)
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2 tháng 1 2022
Áp dụng BĐT Minicopski, ta có:
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2}\\ \Rightarrow P\ge\sqrt{4^2+\left(\dfrac{4}{a+b}\right)^2}=\sqrt{16+\left(\dfrac{4}{4}\right)^2}=\sqrt{17}\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=2\)
15 tháng 8 2021
1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
27 tháng 1 2021
a) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ
nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(A=\dfrac{x-4}{\sqrt{x}+2}\), ta được:
\(A=\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\left(\dfrac{1}{4}-\dfrac{16}{4}\right):\left(\dfrac{1}{2}+2\right)=\dfrac{-15}{4}:\dfrac{5}{2}\)
\(\Leftrightarrow A=\dfrac{-15}{4}\cdot\dfrac{2}{5}=\dfrac{-30}{20}=\dfrac{-3}{2}\)
Vậy: Khi \(x=\dfrac{1}{4}\) thì \(A=\dfrac{-3}{2}\)
b) Ta có: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-1}{2-\sqrt{x}}-\dfrac{9-x}{4-x}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2+x+2\sqrt{x}-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4+9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
27 tháng 1 2021
Thay x = \(\dfrac{1}{4}\)vào bt A ta có: A= \(\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\dfrac{-15}{4}:\dfrac{5}{2}=\dfrac{-3}{2}\)
Vậy x = \(\dfrac{1}{4}\)vào bt A nhận giá trị là -3/2
b)
24 tháng 6 2021
a) đk: x\(\ge0\);
P = \(\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right].\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P = \(\dfrac{8}{9}\)
<=> \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
<=> \(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{3}\)
<=> \(\dfrac{3\sqrt{x}-2x+2\sqrt{x}-2}{3\left(x-\sqrt{x}+1\right)}=0\)
<=> \(-2x+5\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
<=> \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
c)
Đặt \(\sqrt{x}=a\) (\(a\ge0\))
P = \(\dfrac{4a}{3\left(a^2-a+1\right)}\)
Xét P + \(\dfrac{4}{9}\) = \(\dfrac{4a}{3a^2-3a+3}+\dfrac{4}{9}=\dfrac{12a+4a^2-4a+4}{9\left(a^2-a+1\right)}=\dfrac{4a^2+8a+4}{9\left(a^2-a+1\right)}=\dfrac{4\left(a+1\right)^2}{9\left(a^2-a+1\right)}\ge0\)
Dấu "=" <=> a = -1 (loại)
=> Không tìm được Min của P
Xét P - \(\dfrac{4}{3}\) = \(\dfrac{4a}{3\left(a^2-a+1\right)}-\dfrac{4}{3}=\dfrac{4a-4a^2+4a-4}{3\left(a^2-a+1\right)}=\dfrac{-4a^2+8a-4}{3\left(a^2-a+1\right)}=\dfrac{-4\left(a-1\right)^2}{3\left(a^2-a+1\right)}\le0\)
<=> \(P\le\dfrac{4}{3}\)
Dấu "=" <=> a = 1 <=> x = 1 (tm)
PP
2
3 tháng 11 2021
Áp dụng BĐT Bunhiacopski:
\(\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\le\left(1^2+1^2\right)\left(a+1+b+1\right)=2\left(a+b+2\right)\\ \Leftrightarrow a+b+2\ge\dfrac{16}{2}=8\\ \Leftrightarrow a+b\ge6\)
Áp dụng BĐT: \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)
\(\Leftrightarrow P=a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left[\dfrac{\left(a+b\right)^2}{2}\right]^2}{2}=\dfrac{\left(a+b\right)^4}{8}\ge\dfrac{6^4}{8}=162\)
Do đó \(P_{min}=162\Leftrightarrow a=b=3\)
Áp dụng BĐT Min-côp-xki, ta có \(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\sqrt{\left(1+1\right)^2+\left(a^2+b^2\right)^2}=\sqrt{4+\left(a^2+b^2\right)^2}\)
Mà \(\left(a+1\right)\left(b+1\right)=\dfrac{9}{4}\Rightarrow a+b+ab=\dfrac{5}{4}\)
Vì \(ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow a+b+\dfrac{\left(a+b\right)^2}{4}\ge\dfrac{5}{4}\)
\(\Rightarrow4m+m^2-5\ge0\Leftrightarrow\left(m-1\right)\left(m+5\right)\ge0\Rightarrow m\ge1\)(với m=a+b)
\(\Rightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\Rightarrow\left(a^2+b^2\right)^2\ge\dfrac{1}{4}\Rightarrow\sqrt{4+\left(a^2+b^2\right)^2}\ge\dfrac{\sqrt{17}}{2}\)
=> \(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\dfrac{\sqrt{17}}{2}\)