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a)
\(x^2+xy+y^2+1=\left(x^2+2x\times\frac{y}{2}+\left(\frac{y}{2}\right)^2\right)+\frac{3y^2}{4}+1\)
\(=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge0+0+1=1\)
mà\(1>0\Rightarrow x^2+xy+y^2+1>0\)với mọi \(x\)và\(y\)
b)
\(x^2+5y^2+2x-4xy-10y+14\)
\(=\left[x^2+2x\left(1-2y\right)+\left(1-2y\right)^2\right]+y^2-6y+13\)
\(=\left(x+1-2y\right)^2+\left(y^2-2y\times3+9\right)+4\)
\(=\left(x+1-2y\right)^2+\left(y-3\right)^2+4\)
Ta có:\(\left(x+1-2y\right)^2\ge0\)với mọi \(x;y\in R\)
và\(\left(y-3\right)^2\ge0\)với mọi \(x;y\in R\)
\(\Rightarrow\left(x+1-2y\right)^2+\left(y-3\right)^2+4\ge4\)với mọi \(x;y\in R\)
\(\Rightarrow x^2+5y^2+2x-4xy-10y+14>0\)
c)
\(5x^2+10y^2-6xy-4x-2y+3=x^2+4x^2+y^2+9y^2-6xy-4x-2y+3\)
\(=\left[\left(2x\right)^2-2\times2x+1\right]+\left(y^2-2y+1\right)+\left[\left(3y\right)^2-2\times3y+x^2\right]+1\)
\(=\left(2x+1\right)^2+\left(y-1\right)^2+\left(3y-x\right)^2+1\)
Ta có \(\left(2x+1\right)^2\ge0\)với mọi \(x\)
\(\left(y-1\right)^2\ge\)với mọi \(y\)
\(\left(3y-x\right)^2\ge0\)với mọi \(x;y\)
và \(1>0\)
\(\Rightarrow5x^2+10y^2-6xy-4x-2y+3>0\)
a. \(x^2+xy+y^2+1=\left(x^2+xy+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+1=\left(x+\frac{1}{4}y\right)^2+\frac{3}{4}y^2+1>0\forall x;y\)(đpcm)
b. \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left[\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1\right]+\left(y^2-6y+9\right)+4\)
\(=\left[\left(x-2y\right)^2-2\left(x-2y\right)+1\right]+\left(y^2-6y+9\right)+4\)
\(=\left(x-2y-1\right)^2+\left(y-3\right)^2+4>0\forall x;y\)(đpcm)
c. tương tự ý b
_______________Bài làm___________________
a, \(x^2+xy+y^2+1\)
\(=\left(x^2+2x\dfrac{y}{2}+\dfrac{y^2}{4}\right)+\dfrac{3y^2}{4}+1=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^3}{4}+1\)
Do \(\left(x+\dfrac{y}{2}\right)^2\ge0\forall x,y\)
Và \(\dfrac{3y^2}{4}\ge0\forall y\)
Nên: \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0\forall x,y=>đpcm\)
b, \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+\left(y^2-6y+9\right)+5\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+\left(y-3\right)^2+5\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4\)
Do \(\left(x-2y+1\right)^2\ge0\forall x,y\)
Và \(\left(y-3\right)^2\ge0\forall y\)
Nên \(\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)
c, \(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-2x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Do .........
tự làm ik
a/ \(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3}{4}y^2+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge1>0\)
với mọi x,y
b/ \(x^2+5y^2+2x-4xy-16y+14=x^2-2x\left(2y-1\right)+\left(4y^2-4y+1\right)+\left(y^2-12y+36\right)-23\)
\(=\left(x-2y+1\right)^2+\left(y-6\right)^2-23\ge-23\)
Bạn xem lại đề
2 câu trên đã có kết quả, mình giải quyết câu c nhá
5x2 + 10y2 - 6xy - 4x - 2y + 3 > 0
5x2 + 10y2 - 6xy - 4x - 2y + 3 = x2 + 4x2 + y2 + 9y2 - 6xy - 4x - 2y + 3
=[(2x)2 - 2*2x + 1] + (y2 - 2y + 1) + [(3y)2 - 2*3y + x2 ] + 1
=(2x + 1)2 + (y - 1)2 + (3y - x)2 + 1
(2x + 1)2 \(\ge\)0 với mọi x
(y - 1)2 \(\ge\) 0 với mọi y
(3y - x)2\(\ge\) 0 với mọi x và y
1>0
=> ĐPCM
a: \(VT=x^2+2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1>0\forall x,y\)
c: \(VT=x^2-6xy+9y^2+4x^2-4x+1+y^2-2y+1+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x,y\)
a/ \(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)
b/ \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(y^2-6y+9\right)+4\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)
\(5x^2+10y^2-6xy-4x-10y+14\)
\(=\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^2\right)+\left(y^2-10y+25\right)-12\)
\(=\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-5\right)^2-12\ge-12\) đề có nhầm không bạn?