Question

[Phạm Diệu Hằng]

CMR: Các biểu thức sau luôn có giá trị dương với mọi giá trị của các biến

    Q= x² + y² + xy + x + y + 10

 Ai giúp em với ạ, em đang cần rất gấp ạ

Answer 1

[Hoàng Phúc]

\(Q=x^2+y^2+xy+x+y+10\)

\(=\left(x^2+xy+x\right)+y^2+y+10\)

\(=x^2+x\left(y+1\right)+y^2+y+10\)

\(=x^2+2.x.\frac{y+1}{2}+\left(\frac{y+1}{2}\right)^2+y^2+y-\left(\frac{y+1}{2}\right)^2+10\)

\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{\left(y+1\right)^2}{4}+10\)

\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{y^2+2y+1}{4}+10\)

\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{1}{4}y^2-\frac{1}{2}y-\frac{1}{4}+10\)

\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}y^2+\frac{1}{2}y+\frac{39}{4}\)

\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{2}{3}y+13\right)=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+2.y.\frac{2}{6}+\frac{4}{36}-\frac{4}{36}+13\right)\)

\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left[\left(y+\frac{2}{6}\right)^2+\frac{116}{9}\right]=\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\)

Vì \(\left(\frac{2x+y+1}{2}\right)^2\ge0;\frac{3}{4}\left(y+\frac{2}{6}\right)^2\ge0=>\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\ge\frac{29}{3}>0\) (với mọi x;y)

Vậy biểu thức Q luôn dương với mọi giá trị của biến

Answer 2

[✓ ℍɠŞ_ŦƦùM $₦G ✓]

=>4Q=4x²+4xy+4y²+4x+4y+40

=4x²+4x(y+1)+(y+1)²+4y²-y²+4y-2y+40-1

=(2x+y+1)²+3y²+2y+39

\(=\left(2x+y+1\right)^2+\left(\sqrt{3}y+\frac{\sqrt{3}}{3}\right)^2+\frac{116}{3}\)

\(\Rightarrow Q=\left(\frac{2x+y+1}{2}\right)^2+\left(\frac{\sqrt{3}y+\frac{\sqrt{3}}{3}}{2}\right)^2+\frac{29}{3}>0\)

=>đpcm