Cho x, y ∈ R. Chứng minh rằng: 5x2 + xy + 5y2 ≥ \(\frac{11}{4}\)(x + y)2
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1/
a)5x – 20y=5(x-4y)
b) 5x.(x – 1) – 3x(x – 1)=2x(x-1)
c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)
2/
a)x2 + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700
b) x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)2=(53-3)2=2500
3/
a) X + 5x2 = 0
⇒x(x+5)=0
⇒hoặc x=0
x+5=0⇒x=-5
b)x + 1 = ( x + 1 )2
⇒(x + 1)-( x + 1 )2 =0
⇒x(x+1)=0
⇒ hoặc x=0
hoặc x+1=0⇒x=-1
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Vì x;y trái dấu => 2 trường hợp
TH1 y < 0 ; x > 0
TH2 x < 0 ; y > 0
Xét TH1 ta có : \(\frac{xy-x^2}{\sqrt{\frac{-x}{y}}}=\frac{-x\left(x-y\right)}{\sqrt{-\frac{x}{y}}}=\frac{-x\left(x-y\right)}{\sqrt{-\frac{1}{y}}.\sqrt{x}}=\frac{-\left(x-y\right)\sqrt{x}}{\sqrt{-\frac{1}{y}}}=-\left(x-y\right)\left(\sqrt{x.\left(-y\right)}\right)\) ;
\(\frac{xy-y^2}{\sqrt{-\frac{y}{x}}}=\frac{y\left(x-y\right)}{\sqrt{-y}.\sqrt{\frac{1}{x}}}=\frac{-\left(-y\right)\left(x-y\right)}{\sqrt{-y}.\sqrt{\frac{1}{x}}}=-\left(x-y\right)\left(\sqrt{x\left(-y\right)}\right)\)
=> ĐPCM
Xét TH2 ta được \(\frac{xy-x^2}{\sqrt{-\frac{x}{y}}}=\frac{-x\left(x-y\right)}{\sqrt{-x}.\sqrt{\frac{1}{y}}}=\left(x-y\right)\left(\sqrt{-xy}\right)\)
\(\frac{xy-y^2}{\sqrt{\frac{-y}{x}}}=\frac{y\left(x-y\right)}{\sqrt{\frac{1}{-x}}.\sqrt{y}}=\sqrt{-xy}\left(x-y\right)\)
=> ĐPCM
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A = \(\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)= \(\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
= \(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)= \(\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)= \(\frac{2.6}{3.7}=\frac{4}{7}\)
c, theo đề bài ta có :
x2 = yz, y2 = xz , z2 = xy
\(\Rightarrow\frac{x}{y}=\frac{z}{x},\frac{y}{x}=\frac{z}{y},\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)
AD t/c DTSBN, ta có
\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{X+z+y}{y+x+z}=1\)
x= 1y
z= 1x
y= 1z
=> x = y = x
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http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
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Áp dụng BĐT AM-GM ta có: \(xy\le\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\)
Suy ra: \(P=6\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+8\left[\left(x^2+y^2\right)^2-2\left(xy\right)^2\right]+\frac{5}{xy}\)
\(\ge6\left(1-\frac{3}{4}\right)+8\left(\frac{1}{4}-\frac{1}{8}\right)+\frac{5}{\frac{1}{4}}\) (Do x+y=1) \(\Rightarrow P\ge6-\frac{9}{2}+2-1+20=\frac{45}{2}\)(đpcm).
Dấu "=" xảy ra <=> x=y=1/2.
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\(B=2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+2xy+\frac{32}{xy}+\frac{2}{xy}\)
\(B\ge\frac{2.4}{x^2+y^2+2xy}+2\sqrt{2xy.\frac{32}{xy}}+\frac{2}{\frac{\left(x+y\right)^2}{4}}\)
\(B\ge\frac{8}{4^2}+2.8+\frac{8}{4^2}=17\)
Dấu "=" khi \(a=b=2\)
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