Cho X>Y>0 chung minh rang X^3>X^3
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cho 3 so nguyen x,y,z thoa man x+y+z=0 chung minh rang x^3+y^3+z^3= 3xyz
xét hiệu x3+y3+z3-3xyz
=(x+y)3+z3-3xy(x+y)-3xyz
=(x+y+z)3-3(x+y+z)(x+y)z-3xy(x+y+z)
=0 vì x+y+z=0
=>x3+y3+z3=3xyz
=>đpcm
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Cho xy khac 0 va x+y=1
Chung minh rang : x/y^3-1+y/x^3-1-2(xy-2)/(xy)^2+3=0
cho x, y,z >0 chung minh rang\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}< hoac=\frac{3}{ }4\)3/4
cho x,y,z>0 va xyz=1 chung minh rang neu x+y+z>1/x+1/y+1/z thi trong 3 so co it nhat 1 so lon hon 1
cho x/y=y/z/=z/t .chung minh rang:(x+y+z/y+z+t)^3=x/t
Áp dụng tính chất dãy tỉ số bằng ngau ta có :
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}=\dfrac{x+y+z}{y+z+t}\)
\(\Rightarrow\dfrac{x.y.z}{y.z.t}=(\dfrac{x+y+z}{y+z+t})^3\)
\(\Rightarrow\dfrac{x}{t}=(\dfrac{x+y+z}{y+z+t})^3\)
\(\Rightarrowđpcm\)
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cho a=x 3y, b=x 2y 2, c=xy 3 .Chung minh rang voi moi so huu ti x va y ta luon duoc ax+b 2-2x 4y 4=0
cho x, y , z la cac so nguyen thoa man x . y - x. z + y.z - z^2 +1 =0 chung minh rang x+ y =0
Cho x,y duong thoa man: x+y=3. Chung minh rang x2y <= 4
chung minh rang
a) (1-2x)(x-1)-5<0
b) -x^2-y^2+2x+2y-3<0
a) ta có : \(\left(1-2x\right)\left(x-1\right)-5=x-1-2x^2+2x-5\)
\(=-2x^2+3x-6=-\left(2x^2-3x+6\right)=-\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}.\dfrac{3}{2\sqrt{2}}x+\left(\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\right)\)
\(=-\left(\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\right)=-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\)
ta có : \(\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\le0\) với mọi \(x\)
\(-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\le\dfrac{-39}{8}< 0\) với mọi \(x\)
vậy \(\left(1-2x\right)\left(x-1\right)-5< 0\) (đpcm)
b) ta có : \(-x^2-y^2+2x+2y-3\)
\(=\left(-x^2+2x-1\right)+\left(-y^2+2y-1\right)-1\)
\(=-\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-1=-\left(x-1\right)^2-\left(y-1\right)^2-1\)
ta có : \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge\forall x\\\left(y-1\right)^2\ge\forall y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\left(x-1\right)^2\le0\forall x\\-\left(y-1\right)^2\le0\forall y\end{matrix}\right.\)
\(\Rightarrow-\left(x-1\right)^2-\left(y-1\right)^2\le0\) với mọi \(x;y\)
\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2-1\le-1< 0\) với mọi \(x;y\)
vậy \(-x^2-y^2+2x+2y-3< 0\) (đpcm)
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\(a,A=\left(1-2x\right)\left(x-1\right)-5\)
\(=x-1-2x^2+2x-5\)
\(=-2x^2+3x-6\)
\(=-\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{39}{8}\)
\(=-\left[\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right]-\dfrac{39}{8}\)
\(=-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\)
Ta có :
\(-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2\le0\) \(\Rightarrow-\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2-\dfrac{39}{8}\le-\dfrac{39}{8}\)
Hay A \(\le-\dfrac{39}{8}\)
Dấu = xảy ra \(\Leftrightarrow\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2=0\)
\(\Leftrightarrow\sqrt{2}x-\dfrac{3}{2\sqrt{2}}=0\) \(\Leftrightarrow\sqrt{2}x=\dfrac{3}{2\sqrt{2}}\Leftrightarrow x=\dfrac{3}{2\sqrt{2}}:\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
Vậy \(Min_A=-\dfrac{39}{8}\Leftrightarrow x=\dfrac{3}{4}\)
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a) ( 1 - 2x ).( x - 1) -5
= x - 1 - 2x2 + 2x -5
= x - 1 - x2 - x2 + 2x -\(\dfrac{1}{4}+\dfrac{19}{4}\)
= - ( x2 - 2x +1) - [x2 - 2.\(\dfrac{1}{2}\)x + ( \(\dfrac{1}{2}\))2 ] + \(\dfrac{19}{4}\)
= -( x - 1)2 -( x - \(\dfrac{1}{2}\))2 + \(\dfrac{19}{4}\)
Do : -( x - 1)2 =< 0 ; -( x - \(\dfrac{1}{2}\))2 =< 0
--> ( 1 - 2x ).( x - 1) -5 =< -5 < 0 ( ĐPCM)
b) - x2 - y2 + 2x + 2y -3
= - x2 + 2x - 1 - y2 + 2y -1 -1
= - ( x2 - 2x +1) -( y2 - 2y + 1) -1
= -( x - 1)2 - ( y - 1)2 - 1
Do : -( x - 1)2 nhỏ hơn hoặc bằng 0
- ( y - 1)2 nhỏ hơn hoặc bằng 0
--> - x2 - y2 + 2x + 2y -3 =< -5 < 0 ( đpcm)
P/s : Chỗ =< là nhỏ hơn hoặc bằng nhé 
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