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a: \(A=-x^2+4x+5\)
\(=-\left(x^2-4x-5\right)\)
\(=-\left(x^2-4x+4-9\right)\)
\(=-\left(x-2\right)^2+9\le9\)
Dấu '=' xảy ra khi x=2
b: \(B=-4x^2+12x-1\)
\(=-\left(4x^2-12x+1\right)\)
\(=-\left(4x^2-12x+9-8\right)\)
\(=-\left(2x-3\right)^2+8\le8\)
Dấu '=' xảy ra khi x=3/2
\(x^2+4y^2-5x+10y-4xy+20\)
\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)
\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)
\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được :
\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)
\(B=x^2-2xy-2x+2y+y^2\)
\(=x^2-2xy+y^2-2\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được :
\(=1-2=-1\)
\(A=-x^2+2xy-4y^2+x-10y-8\)
=> \(-4A=4x^2-8xy+16y^2-4x+40y+32\)
\(=\left(4x^2-8xy+4y^2\right)-\left(4x-4y\right)+1+12y^2+36y+31\)
\(=\left(2x-2y\right)^2-2\left(2x-2y\right)+1+3\left(4y^2+2.2y.3+9\right)+4\)
\(=\left(2x-2y+1\right)^2+3\left(2y+3\right)^2+4\ge4\)
=> \(A\le4:-4=-1\)
"=" xảy ra <=> \(\hept{\begin{cases}2x-2y+1=0\\2y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-\frac{3}{2}\\x=2\end{cases}}\)
Vậy max A=-1 <=> x=2 y=-3/2
Câu b em làm tương tự nhé!
Bài 1 :
=-5(x^2+4/5x+19/25)
=-5(x^2+2x.2/5+4/25+3/5)
=-5(x+2/5)^2-3
Vì (x+2/5)^2 lớn hơn hoặc bằng 0 =>-5(x+2/5)^2-3 nhỏ hơn hoặc bằng-3
Vậy Min là-3
a) \(C=4x^2+3y^2+4xy-4x-10y+7=\left[4x^2+4x\left(y-1\right)+\left(y-1\right)^2\right]+2\left(y^2-4y+4\right)-2=\left(2x+y-1\right)^2+2\left(y-2\right)^2-2\ge-2\)
\(minC=-2\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=2\end{matrix}\right.\)
d) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
a) 9x2 + y2 + 12x - 10y + 40
= ( 9x2 + 12x + 4 ) + ( y2 - 10y + 25 ) + 11
= ( 3x + 2 )2 + ( y - 5 )2 + 11 ≥ 11 ∀ x, y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+2=0\\y-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=5\end{cases}}\)
Vậy GTNN của biểu thức = 11 <=> x = -2/3 ; y = 5
b) 2x2 + 2y2 - 4x - 4y - 2xy + 30
= ( x2 - 2xy + y2 ) + ( x2 - 4x + 4 ) + ( y2 - 4y + 4 ) + 22
= ( x - y )2 + ( x - 2 )2 + ( y - 2 )2 + 22 ≥ 22 ∀ x, y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)
Vậy GTNN của biểu thức = 22 <=> x = y = 2
a) Đặt \(A=9x^2+y^2+12x-10y+40\)
\(\Rightarrow A=\left(9x^2+12x+4\right)+\left(y^2-10y+25\right)+11\)
\(=\left(3x+2\right)^2+\left(y-5\right)^2+11\)
Vì \(\left(3x+2\right)^2\ge0\forall x\); \(\left(y-5\right)^2\ge0\forall y\)
\(\Rightarrow\left(3x+2\right)^2+\left(y-5\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(3x+2\right)^2+\left(y-5\right)^2+11\ge11\forall x,y\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3x+2=0\\y-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-2\\y=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{3}\\y=5\end{cases}}\)
Vậy \(minA=11\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{3}\\y=5\end{cases}}\)
b) Đặt \(B=2x^2+2y^2-4x-4y-2xy+30\)
\(\Rightarrow B=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+22\)
\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2+22\)
Vì \(\left(x-y\right)^2\ge0\forall x,y\); \(\left(x-2\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2+22\ge22\forall x,y\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\x=2\\y=2\end{cases}}\Leftrightarrow x=y=2\)
Vậy \(minB=22\)\(\Leftrightarrow x=y=2\)