Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
Xét hiệu:
$x^2+4y^2+9-2xy-3x-6y$
$=\frac{1}{2}(x^2+4y^2-4xy)+\frac{1}{2}(x^2-6x+9)+\frac{1}{2}(4y^2-12y+9)$
$=\frac{1}{2}(x-2y)^2+\frac{1}{2}(x-3)^2+\frac{1}{2}(2y-3)^2$
$\geq 0$ với mọi $x,y\in\mathbb{R}$
$\Rightarrow x^2+4y^2+9\geq 2xy+3x+6y$
Ta có đpcm
Dấu "=" xảy ra khi $x=3; y=\frac{3}{2}$
\(Tacó\): \(C=x^2+2xy+y^2+y^2-6y+15\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+6\)
\(=\left(x+y\right)^2+\left(y-3\right)^2+6\)
\(Mà\)\(\left(x+y\right)^2\ge0\)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
a)\(x^2+4y^2-2x+4y+2\)
\(=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2\ge0\)(đúng)
b) Sửa đề
\(3y^2+x^2+2xy+2x+6y+3\)
\(=\left(x^2+y^2+2xy\right)+2y^2+2x+6y+3\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+2y^2+4y+2\)
\(=\left(x+y+1\right)^2+2\left(y+1\right)^2\ge0\) (đúng)
Bài 1:
Ta có:
\(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Ta có:
\(-\left(4x-x^2-5\right)=-4x+x^2+5=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\ge1>0\)
\(\Rightarrow4x-x^2-5< 0\)
\(\left(3x-6y\right)\left(x^2+2xy+4y^2\right)-3\left(x^3-8y^3+12\right)\)
\(=3\left(x-2y\right)\left(x^2+2xy+4y^2\right)-3\left(x^3-8y^3+12\right)\)
\(=3\left(x^3-8y^3\right)-3\left(x^3-8y^3+12\right)\)
=-36
