a) x²-6x+10>0

<=>x²-6x+9+1>0

<=>(x-3)²+1>0(đúng với mọi x)

vậy x²-6x+10>0 với mọi x

b)x²-2x+y²+4y+6>0 

<=>x²-2x+1y²+4y+4+1>0

<=>(x-1)²+(y+2)²+1>0 (với mọi x,y)

Vậy x²-2x+y²+4y+6>0 với mọi x,y

\(Tacó\):   \(C=x^2+2xy+y^2+y^2-6y+15\)

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

\(=\left(x+y\right)^2+\left(y-3\right)^2+6\)

\(Mà\)\(\left(x+y\right)^2\ge0\)với mọi x,y

             \(\left(y-3\right)^2\ge0\)với mọi y

\(\Rightarrow\left(x+y\right)^2+\left(y-3\right)^2+6>0\)

\(Hay\)\(x^2+2xy+y^2+y^2-6y+15>0\)\

: 

Ta có C = (x² + 2xy + y²) + (y² - 6x + 9) + 6 

= (x + y)² + (y - 3)² + 6 \(\ge6>0\)(đpcm)

C = x² + 2xy + y² + y² - 6y + 15 

C = ( x² + 2xy + y² ) + ( y² - 6y + 9 ) + 6

C = ( x + y )² + ( y - 3 )² + 6 ≥ 6 > 0 ∀ x ( đpcm )

D = x² + y² + 6x + 10y + 30

D = ( x² + 6x + 9 ) + ( y² + 10y + 25 ) - 4

D = ( x + 3 )² + ( y + 5 )² - 4 ≥ -4 ( xem lại đề nhớ )

\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)

ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)

Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)

T i c k cho mình 1 cái nha mới bị trừ 50 đ

Ta có: \(2x^2+4y^2+4xy-6x+10\)\(=x^2+4xy+4y^2+x^2-6x+9+1\)\(=\left(x+2y\right)^2+\left(x-3\right)^2+1\)

Vì \(\left(x+2y\right)^2\ge0;\left(x-3\right)^2\ge0\)\(\Rightarrow\left(x+2y\right)^2+\left(x-3\right)^2\ge0\)\(\Leftrightarrow\left(x+2y\right)^2+\left(x-3\right)^2+1\ge1>0\)\(2x^2+4y^2+4xy-6x+10>0\left(đpcm\right)\)

Câu hỏi của KiKyo - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

       \(x^2+4y^2+z^2-2x-6z+8y+15\)

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

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

       \(x^2+5y^2+2x-4xy-10y+14\)

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

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

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

Chúc bạn học tốt.

bài 1 phân tích da thức hả bạn

\(=x^2+4y^2+4xy+x^2-6x+9+1=\left(x+2y\right)^2+\left(x-3\right)^2+1\)

Ta có: \(\left(x+2y\right)^2\ge0;\left(x-3\right)^2\ge0\left(\forall x;y\right)\)

\(\Rightarrow\left(x+2y\right)^2+\left(x-3\right)^2+1\ge1>0\forall x;y\)

=> đpcm