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\(A=3x^2+5x-2\)
\(A=3\left(x^2+\frac{5}{3}x-\frac{2}{3}\right)\)
\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2-\frac{49}{36}\right)\)
\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2\right)-\frac{49}{12}\)
\(A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\)
Vì \(3\left(x+\frac{5}{6}\right)^2\ge0\)
Do đó \(3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)
Dấu = xảy ra khi \(x+\frac{5}{6}=0\Rightarrow x=-\frac{5}{6}\)
Vậy Min A=\(-\frac{49}{12}\) khi x=\(-\frac{5}{6}\)
mk làm ý a thôi, mấy ý sau dựa vào mà làm.
A = \(3x^2+5x-2\)
=> \(\frac{A}{3}=x^2+\frac{5}{3}x-\frac{2}{3}\)(chia cả 2 vế cho 3)
\(\Leftrightarrow\frac{A}{3}=x^2+2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2-\frac{49}{36}\)
\(\Leftrightarrow\frac{A}{3}=\left(x+\frac{5}{6}\right)^2-\frac{49}{36}\)
\(\Rightarrow A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)
Đẳng thức xảy ra <=> x = - 5/6.
Vậy Min A = - 49/12 khi và chỉ khi x = - 5/6.
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
D ez nhất :v
\(D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+5\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+5\ge5\)
Đẳng thức xảy ra khi x = 1 và y = -2
\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)
\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)
\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)
Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1
a, \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
Thay x + y = 3
\(\Leftrightarrow A=9-12+1=-2\)
Vậy A = -2 khi x + y = 3
b, \(B=x^2+4y^2-2x+10+4xy-4y\)
\(=x^2+4xy+4y^2-2x-4y+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 có:
\(B=25-10+10=25\)
Vậy B = 25 khi x + 2y = 5
1. A = x2 + 2y2 + 2xy - 4y - 3
= ( x2 + 2xy + y2 ) + ( y2 - 4y + 4 ) - 7
= ( x + y )2 + ( y - 2 )2 - 7
Vì ( x + y )2\(\ge\)0\(\forall\)x;y ; ( y - 2 )2\(\ge\)0\(\forall\)y
=> A = ( x + y )2 + ( y - 2 )2 - 7 \(\ge\)- 7
Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-2\\y=2\end{cases}}\)
Vậy minA = - 7 <=> \(\orbr{\begin{cases}x=-2\\y=2\end{cases}}\)
2. B = 3x2 + 4y2 + 4xy - 4x - 2
= ( x2 + 4xy + 4y2 ) + ( 2x2 - 4x + 2 ) - 4
= ( x + 2y )2 + 2 ( x - 1 )2 - 4\(\ge\)- 4
Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(x+2y\right)^2=0\\2\left(x-1\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}y=-\frac{x}{2}\\x=1\end{cases}}\)<=> \(\orbr{\begin{cases}y=-\frac{1}{2}\\x=1\end{cases}}\)
Vậy minB = - 4 <=> \(\orbr{\begin{cases}y=-\frac{1}{2}\\x=1\end{cases}}\)
C = 4x2 + 2y2 - 4xy - 6y + 5
= ( 4x2 - 4xy + y2 ) + ( y2 - 6y + 9 ) - 4
= ( 2x - y )2 + ( y - 3 )2 - 4\(\ge\)- 4
Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(2x-y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{y}{2}\\y=3\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{3}{2}\\y=3\end{cases}}\)
Vậy minC = - 4 <=> \(\orbr{\begin{cases}x=\frac{3}{2}\\y=3\end{cases}}\)