Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có: \(\left(a-\frac{1}{2}\right)^2\ge0\Leftrightarrow a^2-a+\frac{1}{4}\ge0\Leftrightarrow a^2+\frac{1}{4}\ge a\)
Tương tự cũng có : \(b^2+\frac{1}{4}\ge b ; c^2+\frac{1}{4}\ge c\)
Cộng vế với vế các bất đẳng thức cùng chiều ta đươc:
\(a^2+b^2+c^2+\frac{3}{4}\ge a+b+c\)( Vì a + b + c = \(\frac{3}{2}\) nên \(a^2+b^2+c^2\ge\frac{3}{4}\))
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{2}\)
a)\(x^2-4x+y^2-2y+10=\left(x^2-4x+4\right)+\left(y^2-2y+1\right)+5\)
\(=\left(x-2\right)^2+\left(y-1\right)^2+5\ge5\)
Dấu "=" xảy ra khi x=2;y=1
b) tương tự câu a
c)\(x^2+2y^2-6x-8y+2xy+5=x^2+2y^2+2x\left(y-3\right)-8y+5\)
\(=x^2+2x\left(y-3\right)+\left(y^2-6x+9\right)+\left(y^2-2x+1\right)-5\)
\(=x^2+2x\left(y-3\right)+\left(y-3\right)^2+\left(y-1\right)^2-5\)
\(=\left(x+y-3\right)^2+\left(y-1\right)^2-5\ge-5\)
Dấu "=" xảy ra khi x=2;y=1
Giải:
\(P=x^2+2y^2+2xy-6x-8y+2018\)
\(\Leftrightarrow P=\left(x^2+y^2+9+2xy-6x-6x\right)+\left(y^2-2y+1\right)+2008\)
\(\Leftrightarrow P=\left(x+y-3\right)^2+\left(y-1\right)^2+2008\)
Vì \(\left\{{}\begin{matrix}\left(x+y-3\right)^2\ge0;\forall x,y\\\left(y-1\right)^2\ge0;\forall y\end{matrix}\right.\)
\(\Leftrightarrow\left(x+y-3\right)^2+\left(y-1\right)^2+2008\ge2008;\forall x,y\)
Hay \(P\ge2008;\forall x,y\)
Vậy ...
\(P=x^2+2y^2+2xy-6x-8y+2018\)
<=> \(P=\left(x^2+2xy+y^2\right)-\left(6x+6y\right)+9+\left(y^2-2y+1\right)+2008\)
<=> P=(x+y)2-6(x+y) +9 +(y-1)2 +2008
<=> P=(x+y-3)2+(y-1)2+2008
=> Min P= 2008 dấu = xảy ra khi y=1;x=2
Ta có
\(A=x^2+2y^2+2xy-2x-8y+2017\)
\(=\left(x^2+2xy+y^2\right)-2\left(x+y\right)+1+\left(y^2-6y+9\right)+2007\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1+\left(y-3\right)^2+2007\)
\(=\left(x+y-1\right)^2+\left(y-3\right)^2+2007\ge2007\)
Dấu = xảy ra khi \(\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
1a) A = \(x^2-4x+2023=\left(x-2\right)^2+2019\)
Ta luôn có: (x - 2)2 \(\ge\)0 \(\forall\)x
=> (x - 2)2 + 2019 \(\ge\)2019 \(\forall\)x
Hay A \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra khi : (x - 2)2 = 0 => x - 2 = 0 => x = 2
Nên Amin = 2019 khi x = 2
a) A=x2-3x+1
A= x2-2.x.\(\dfrac{3}{2}\) +\(\dfrac{9}{4}\) +1-\(\dfrac{9}{4}\)
A=(x2-2.x.\(\dfrac{3}{2}\) +\(\dfrac{9}{4}\) )-\(\dfrac{5}{4}\)
A=(x-\(\dfrac{3}{2}\) )2- \(\dfrac{5}{4}\)
\(Do\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\)
=>\(\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
=>A\(\ge-\dfrac{5}{4}\)
vậy GTNN của A= -\(\dfrac{5}{4}\) khi
\(x-\dfrac{5}{4}=0\)
<=>x=\(\dfrac{5}{4}\)
\(P=x^2+\left(2xy-6x\right)+2y^2-8y+2029\)
\(P=x^2+2x\left(y-3\right)+\left(y-3\right)^2-\left(y-3\right)^2+2y^2-8y+2029\)
\(P=\left(x+y-3\right)^2-\left(y^2-6y+9\right)+2y^2-8y+2029\)
\(P=\left(x+y-3\right)^2+y^2-2y+1+2019\)
\(P=\left(x+y-3\right)^2+\left(y-1\right)^2+2019\) \(\ge2019\forall x,y\)
\(P=2019\Leftrightarrow\left\{{}\begin{matrix}x+y-3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy Min P = 2019 \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
1.\(\Leftrightarrow a^2+b^2-ab-a-b+3>0\)
\(\Leftrightarrow2a^2+2b^2-2ab-2a-2b+6>0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+4>0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+4>0\) ( luôn đúng )
Do đó suy ra đpcm