cho a>b>0 và ab=1
P=\(\dfrac{a\sqrt{b}+b\sqrt{a}+a\sqrt{a}+b\sqrt{b}}{a-b}\)
tìm Pmin
rất khó nha
K
Khách
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30 tháng 8 2022
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
NS
17 tháng 7 2018
Áp dụng bất đẳng thức Mincopxki:
\(\sqrt{a^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{b^2}+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{c^2}+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(=\sqrt{\left(a+b+c\right)^2+2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+2.\left(\dfrac{9}{a+b+c}\right)^2}\) ( Cauchy-Schwarz)
\(=\sqrt{\left(a+b+c\right)^2+\dfrac{162}{\left(a+b+c\right)^2}}=\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
\("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
20 tháng 6 2018
Câu a
\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\left(\sqrt{a}+\sqrt{b}\right):\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{1}\)
\(=a-b\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
14 tháng 8 2023
a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
26 tháng 5 2018
Áp dụng bđt AM-GM: \(\dfrac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\dfrac{2\sqrt{ab}}{2\sqrt{\sqrt{ab}}}=\sqrt{\sqrt{ab}}\)
4 tháng 7 2017
VT= \(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b^2}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{a^2}-\sqrt{ab}+\sqrt{b^2}-\sqrt{ab}\)\(=\sqrt{a^2}-2\sqrt{ab}+\sqrt{b^2}=\left(\sqrt{a}-\sqrt{b}\right)^2\)=VP
=> đpcm.
\(P=\dfrac{\sqrt{a.b}\left(\sqrt{a}+\sqrt{b}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(P=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(P=\dfrac{a+b}{\sqrt{a}-\sqrt{b}}\)
tới đây là tự tìm đc rồi
Lời giải:
Ta có: \(a\sqrt{b}+b\sqrt{a}+a\sqrt{a}+b\sqrt{b}=(a\sqrt{b}+b\sqrt{a})+(a\sqrt{a}+b\sqrt{b})\)
\(=\sqrt{ab}(\sqrt{a}+\sqrt{b})+(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)\)
\(=(\sqrt{a}+\sqrt{b})(\sqrt{ab}+a-\sqrt{ab}+b)=(\sqrt{a}+\sqrt{b})(a+b)\)
Và: \(a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\)
Do đó: \(P=\frac{a+b}{\sqrt{a}-\sqrt{b}}=\frac{a+b-2\sqrt{ab}+2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{(\sqrt{a}-\sqrt{b})^2+2}{\sqrt{a}-\sqrt{b}}=(\sqrt{a}-\sqrt{b})+\frac{2}{\sqrt{a}-\sqrt{b}}\)
\(\geq 2\sqrt{(\sqrt{a}-\sqrt{b}).\frac{2}{\sqrt{a}-\sqrt{b}}}=2\sqrt{2}\) (áp dụng BĐT Cô-si cho 2 số dương)
Vậy \(P_{\min}=2\sqrt{2}\)
Dấu "=" xảy ra khi \((a,b)=(2+\sqrt{3}, 2-\sqrt{3})\)