Còn ai không ạ :(
Chứng minh rằng không tồn tại các số thực x, y, z thỏa mãn:
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3 = 0
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15 = 0
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a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1
Ta có : \(\hept{\begin{cases}\left(x-3y\right)^2\\\left(2x-1\right)^2\\\left(y-1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1>0\forall x,y\)
=> đpcm
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15 = 0 < Sửa -z2 -> +z2 )
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + 4( y2 + 2y + 1 ) + ( z - 3 )2 + 1
= ( x - 1 )2 + 4( y + 1 )2 + ( z - 3 )2 + 1
Ta có : \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\4\left(y+1\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\Rightarrow\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge1>0\forall x,y,z\)
=> đpcm
\(14,P=x^2+xy+y^2-3x-3y+3\\ P=\left(x^2+xy+\dfrac{1}{4}y^2\right)-3\left(x+\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2-\dfrac{3}{2}y+3\\ P=\left(x+\dfrac{1}{2}y\right)^2-3\left(x+\dfrac{1}{2}y\right)+\dfrac{9}{4}+\dfrac{3}{4}\left(y^2-2y+1\right)\\ P=\left(x+\dfrac{1}{2}y-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2\ge0\)
\(x^2+4y^2+z^2-2x+8y-6x+15=0\)
<=> \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1=0\)
mà \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\)≥0
=> \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)≥1
=> ko có giá trị nào của x,y,z thỏa mãn
\(A=\dfrac{1}{x^2-4x+9}=\dfrac{1}{\left(x-2\right)^2+5}\)
mà (x+2)2≥0
=> (x+2)2+5≥5
=> \(\dfrac{1}{\left(x-2\right)^2+5}\)≤ 1/5
=> Max A = 1/5 dấu ''='' xảy ra khi x=2
⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0
x^2+4y^2+z^2-2x-6z+8y+15
=x^2+4y^2+z^2-2x-6z+8y+1+1+4+9
=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1
=(x-1)^2+4(y+1)^2+(z-3^)2+1
Ta thấy:(x−1)^2≥0
4(y+1)^2≥0
(z−3)^ 2≥0
{(x−1)^24(y+1)^2(z−3)^2≥0
⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0
⇒(x−1)2+4(y+1)2+(z−3)2+1≥0+1=1>0
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + ( 2y + 2 )2 + ( z - 3 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm