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Áp dụng bất đẳng thức cô si ta có :
\(x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3+\left(\sqrt{z}\right)^3\ge3\sqrt[3]{\left(\sqrt{xyz}\right)^3}=3\sqrt{xyz}\)Dấu "=" xảy ra khi :\(\sqrt{x}=\sqrt{y}=\sqrt{z}\)
Do đó :\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Vậy A=8
bài 3:
a, đặt x12=y9=z5=kx12=y9=z5=k
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: x5=y7=z3=x225=y249=z29x5=y7=z3=x225=y249=z29
A/D tính chất dãy tỉ số bằng nhau ta có:
x5=y7=z3=x225=y249=z29=x2+y2−
\(\sqrt{x}+\sqrt{y}=\sqrt{z}\Rightarrow x+y+2\sqrt{xy}=z\Rightarrow x+y-z=-2\sqrt{xy}\)
\(\sqrt{x}-\sqrt{z}=\sqrt{y}\Rightarrow x+z-2\sqrt{xz}=y\Rightarrow z+x-y=2\sqrt{xz}\)
Tương tự:\(y+z-x=2\sqrt{yz}\)
\(A=\frac{1}{-2\sqrt{xy}}+\frac{1}{2\sqrt{yz}}+\frac{1}{2\sqrt{zx}}=\frac{1}{2}\left(\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\sqrt{xyz}}\right)=0\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
Ta có : \(3\sqrt{xyz}=\sqrt{x}^2+\sqrt{y}^3+\sqrt{z}^3\ge3\sqrt[3]{\sqrt{x}^3\sqrt{y}^3\sqrt{z}^3}=3\sqrt{x}\sqrt{y}\sqrt{z}=3\sqrt{xyz}.\)
Dấu = xảy ra
=> x =y =z
=> A = (1+1)(1+1)(1+1) =8
[Verse 1: Rap Monster]
Ai tai
Kotoba ni suru to yokei
Ai tai
Kimi no shashin miru to ne
Ai tai demo jikan ga nai
Tada sugi te iku
Toki no naka mizukara o seme mata
Tōku naru negai
[Verse 2: Rap Monster, Jungkook]
Marude fuyu no yō sa natsu de mo fubuku yō sa
Kokoro o nose ta ressha sudeni yuki no naka
Isso kimi to chikyū no uragawa e te o tsukan de mō nige tai
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Friend
[Verse 3: Jimin, V]
Kono ōzora o mau koyuki no yō
Koyuki no yō
Maeru nara kimi e
Sugu tadoritsukeru hazu na noni
[Pre-Chorus: Jungkook, J-Hope, V, Jin]
Yuki no hanabira
Maiori
Kie te itta
Ai tai?
Ai tai?
Dore hodo omoe ba
Kogoeru yoru kazoere ba
Nē, aeru no?
Deaeru no?
[Chorus: Jungkook, Jin, J-Hope]
Fuyu no owari o tsugeru
Yasashii haru no hi made
Hana saku made
I te hoshii sonomama
Sonomama
[Verse 4: SUGA, Jimin]
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Ja nakya boku ga kawatta
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Tsumoru omoide wa konnani mada nokotte iru noni
[Pre-Chorus: Jungkook, V, J-Hope, Jin]
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Donna yoru mo donna kisetsu mo
Owari wa kuru kara
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Fuyu ga mō owaru yo
Ai tai
Ai tai
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Lời giải:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{x+y-z}\Leftrightarrow \frac{x+y}{xy}=\frac{1}{z}+\frac{1}{x+y-z}=\frac{x+y}{z(x+y-z)}\)
\(\Leftrightarrow (x+y)(\frac{1}{xy}-\frac{1}{z(x+y-z)})=0\)
\(\Leftrightarrow (x+y).\frac{z(x+y-z)-xy}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y).\frac{(z-x)(y-z)}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y)(z-x)(y-z)=0\)
Xét các TH sau:
TH1: $x+y=0$. TH này loại do ĐKXĐ $x,y>0$
TH2: $z-x=0\Leftrightarrow z=x$
$\Leftrightarrow \frac{1}{y}=\frac{2020}{2021}$
\(M=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{y}}=2\sqrt{\frac{2020}{2021}}\)
TH3: $y-z=0$ tương tự TH2, ta có \(M=2\sqrt{\frac{2020}{2021}}\)