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a) \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+\left(y^2-6y+9\right)+4\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)

\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4\)

Vì \(\left(x-2y+1\right)^2\ge0\)

      \(\left(y-3\right)^2\ge0\)

 \(\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2+4\ge4>0\)với mọi x,y (ĐPCM)
b) \(5x^2+10y^2-6xy-4x-2y+3\)

\(=\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^2\right)+\left(y^2-2y+1\right)+1\)

\(=\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-1\right)^2+1\)

Vì \(\left(2x-1\right)^2\ge0\)

      \(\left(x-3y\right)^2\ge0\)

       \(\left(y-1\right)^2\ge0\)

 \(\Rightarrow\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-1\right)^2+1\ge1>0\)vợi mọi x,y (ĐPCM)

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

Các bạn giúp mình vs

bạn làm rõ số mũ ở đâu ra dùm mình nhé, mình giải hết cho, nhưng câu b sai đề nhé bạn

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: \(VT=x^2-6xy+9y^2+4x^2-4x+1+y^2-2y+1+1\)

\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x,y\)