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x+y=2
\(\Rightarrow\)x=1; x=0; x=-1; x=-2;...
y=1; y=2; y=3; y=4;...
\(\Rightarrow\)x.y= 1.1=1=1
0.2=0<1
-1.3=-3<1
-2.4=-8<1
.............
\(\Rightarrow\)Nếu x+y=2 thì x.y\(\le\)1
Ta có: \(x+y=2\)
\(\Rightarrow x=2-y.\)
Có: \(x.y=\left(2-y\right).y\)
\(\Rightarrow x.y=2y-y^2\)
\(\Rightarrow x.y=-y^2+2y-1+1\)
\(\Rightarrow x.y=-\left(y-1\right)^2+1.\)
Vì \(\left(y-1\right)^2\ge0\) \(\forall y.\)
\(\Rightarrow-\left(y-1\right)^2\le0\) \(\forall y.\)
\(\Rightarrow-\left(y-1\right)^2+1\le1\) \(\forall y.\)
\(\Rightarrow x.y\le1\left(đpcm\right).\)
Chúc bạn học tốt!
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x+y=2
=> (x+y)2=4
=> x2+y2+2xy = 4
Áp dụng x2+y2 >= 2xy
=> x2+y2+2xy >= 4xy
Mà x2+y2+2xy = 4
=> 4>= 4xy
=> xy <= 1
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Ta có: x+y=2⇒y=2−x
Khi đó:x.y=x(2−x)=2x−x2
=1−(x2−2x+1)
=1−(x−1)2≤1
=>x.y≤1(đpcm)
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\(\left(2a-3\right)\left(\frac{3}{4}a+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2a-3=0\\\frac{3}{4}a+1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2a=3\\\frac{3}{4}a=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}a=\frac{3}{2}\\a=-\frac{4}{3}\end{array}\right.\)
\(\left(2a-3\right)\left(\frac{3}{4}a+1\right)=0\)
<=> \(\left[\begin{array}{nghiempt}2a-3=0\\\frac{3}{4}a+1=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}a=\frac{3}{2}\\a=-\frac{4}{3}\end{array}\right.\)
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Ta có \(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)\(\Rightarrow x-2\sqrt{xy}+y\ge0\)\(\Rightarrow x+y\ge2\sqrt{xy}\)
Mà x + y = 2 \(\Rightarrow\)\(2\ge2\sqrt{xy}\)\(\Rightarrow1\sqrt{xy}\le1\)\(\Rightarrow xy\le1\)
Vi 2 = 2 + 0 ; 1 + 1 .nen x.y = 2 . 0 ; 1.1 chi bang 0 hoac 1 nen x.y <= 1