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\(A=\left(x+1\right)^2-\left(2x-3\right)^2-15\)
\(A=\left(x+1\right)^2-\left(2x-3\right)^2-15\ge-15\)
\(A_{min}=-15\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}}\)
P/s tham khảo nha
\(B=x^2-2xy+4y^2-2x-10y+2018\)
\(B=\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+\left(3y^2-12y+12\right)+2005\)
\(B=\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2005\)
\(B=\left(x-y-1\right)^2+3\left(y-2\right)^2+2005\ge2005\)
VÌ \(\left(x-y-1\right)^2+3\left(y-2\right)^2\ge0\forall x;y\)
DẤU "="XẢY RA KHI Y=2;X=3
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A=1/2017-2/2017x+2018/2017x^2
1/2017x=y
A=2018y^2-2y+1/2017
A=2018(y^2-2.y./2018+1/2018^2)
-1/2018+1/(2017)
A=2018(y-1/2018)^2-1/(2018.2017)
GTNNA=1/(2017.2018)
khi y=1/2018
x=2017/2018
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a) \(H=x^2-4x+16\)
\(H=\left(x+2\right)^2+12\ge12\)
vậy min H=12 \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
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Đặt biểu thức là A
\(x^2+xy+y^2-3x-3y+2018\)
\(=\left(x^2+xy+y^2\right)-\left(3x+3y\right)+2018\)
\(=\left(x+y\right)^2-3\left(x+y\right)+2018\)
Ta có : (x - y)² ≥ 0
<=> x² + y² ≥ 2xy
<=> x² + 2xy + y² ≥ 4xy
<=> (x + y)² ≥ 4xy
<=> xy ≤ (x + y)²/4
<=> -xy ≥ -(x + y)²/4
--> A ≥ (x + y)² - 3(x + y) - (x + y)²/4
<=> A ≥ 3(x + y)²/4 - 3(x + y)
để dễ nhìn,ta đặt t = x + y
--> A ≥ 3t²/4 - 3t = 3(t²/4 - 2.t/2 + 1) - 3 = 3(t/2 - 1)² - 3 ≥ -3
Dấu " = " xảy ra <=> t/2 = 1 <=> t = 2 <=> x + y = 2 và x = y --> x = y = 1
Vậy MinA = -3 <=> x = y = 1
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Bài 2:
a: \(=-\left(x^2+4x-10\right)\)
\(=-\left(x^2+4x+4-14\right)=-\left(x+2\right)^2+14< =14\)
Dấu = xảy ra khi x=-2
b: \(=-2\left(x^2-2x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-2x+1+\dfrac{3}{2}\right)\)
\(=-2\left(x-1\right)^2-3< =-3\)
Dấu = xảy ra khi x=1
c: \(=x^2-2x+1-2\left(x^2+6x+9\right)+20\)
\(=x^2-2x+21-2x^2-12x-18\)
\(=-x^2-14x+3\)
\(=-\left(x^2+14x-3\right)\)
\(=-\left(x^2+14x+49-52\right)=-\left(x+7\right)^2+52< =52\)
Dấu = xảy ra khi x=-7
Ta có \(\frac{x^2-2x+2018}{x^2}=1-\frac{2}{x}+\frac{2018}{x^2}=2018\left(\frac{1}{x^2}-\frac{2}{2018x}+\frac{1}{2018}\right)=2018\left(\frac{1}{x^2}-2.\frac{1}{2018x}+\frac{1}{2018^2}\right)+\frac{2017}{2018}=2018.\left(\frac{1}{x}-\frac{1}{2018}\right)^2+\frac{2017}{2018}\)
Nhận thấy \(2018\left(\frac{1}{x}-\frac{1}{2018}\right)^2\ge0\forall x=>2018\left(\frac{1}{x}-\frac{1}{2018}\right)^2+\frac{2017}{2018}\ge\frac{2017}{2018}\forall x\)
Dấu "=" xảy ra khi 1/x-1/2018=0=> x=2018
Vậy min A=2017/2018 <=> x=2018