3) 5x² + y² -4xy - 2y + 8x + 2013

= ( 4x² + y² -4xy -2y + 8x ) + x² + 2013

= ( 2x - y +1)² + x² +2013

Vì ( 2x-y+1)² \(\ge\)0 \(\forall x,y\); x² \(\ge\)0\(\forall x\)

=> (2x - y+1)² + x² \(\ge\)0

=> ( 2x-y +1)² +x² + 2013\(\ge\)0

hay  A \(\ge0\)\(\forall x,y\)=> A ko âm

Giúp mk phần 1 và phần 2 vs!!!

HELP ME PLEASE!!!

1\(\left(15x-1\right)^2+3\left(7x+3\right)\left(x+1\right)-\left(x^2-73\right)\))

\(=\left(15x-1\right)^2+21x^2+30x+9-x^2+73\)

\(=\left(15x-1\right)^2+20x^2+30x+82\)

\(=\left(15-1\right)^2+20\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+\frac{283}{4}\)

\(=\left(15x-1\right)^2+20\left(x+\frac{3}{4}\right)^2+\frac{283}{4}\)

Vì \(\left(15x-1\right)^2;20\left(x+\frac{3}{4}\right)^2;\frac{283}{4}\ge0\forall x\)=> Biểu thức ko âm

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

\(M=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)

\(M=\left(x+3y\right)^2+\left(2x+1\right)^2+\left(y-4\right)^2+1>0;\forall x;y\)

Ta có: \(M=18+4x-8y+6xy+5x^2+10y^2\)

\(=4x^2+4x+1+x^2+6xy+9y^2+y^2-8y+16+1\)

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

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\left(x+3y\right)^2\ge0\forall x,y\)

\(\left(y-4\right)^2\ge0\forall y\)

Do đó: \(\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\ge1>0\forall x,y\)

hay \(M>0\forall x,y\)

a, \(x^2+xy+y^2+1=x^2+\dfrac{1}{2}xy+\dfrac{1}{2}xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)

\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\)

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left(x^2+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)

\(\Rightarrow\left(x^2+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\ge1\)

Vậy............

b, \(5x^2+10y^2-6xy-4x-2y+3\)

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

\(=x^2-3xy-3xy+9y^2+4x^2-2x-2x+1+y^2-y-y+1+1\)

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

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

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2\ge0\)

\(\Rightarrow\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)

Vậy..............

Chúc bạn học tốt!!!

_______________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

\(M=18+4x-8y+6xy+5x^2+10y^2\)

\(=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)

\(=\left(x+y\right)^2+4\left(x+\frac{1}{2}\right)^2+\left(y-4\right)^2+1\)

Có \(\left(x+y\right)^2\ge0\forall xy\)

\(4\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\left(y-4\right)^2\ge0\forall y\)

\(\Rightarrow M\ge1\forall x,y\)

hay \(M>0\forall x,y\)

Mấy bạn bị lms í=)) dễ v cũng ko biết làm

Mình chỉ đăng lên để thử xem coi ai làm đc ko chứ mình cx ko biết làm. Ai jup mình vớiiiiii