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\(A=\frac{3}{\left(x+2\right)^2+4};\left(x+2\right)^2\in N\)
\(\Rightarrow A_{max}\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2+4=4\)
\(\Rightarrow A_{max}=\frac{3}{4}\)
b, \(B=\left(x+1\right)^2+\left(y+3\right)^2+1\)
Mặt khác: \(\left(x+1\right)^2;\left(y+3\right)^2\in N\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\)
\(\Rightarrow B_{min}\Leftrightarrow\left(x+1\right)^2+\left(y+3\right)^2=0\Rightarrow B_{min}=1\)
\(A=\frac{3}{\left(x+2\right)^2+4}\)
Để A max
=>(x+2)^2+4 min
Mà\(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+4\ge4\)
Vậy Min = 4 <=>x=-2
Vậy Max A = 3/4 <=> x=-2
\(b,B=\left(x+1\right)^2+\left(y+3\right)^2+1\)
Có \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\)
\(\Rightarrow B\ge0+0+1=1\)
Vậy MinB = 1<=>x=-1;y=-3
Bài 2:
a) \(A=x^2+6\ge6>0\forall x\in R\)
b) \(B=\left(5-x\right)\left(x+8\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5-x>0\\x+8>0\end{matrix}\right.\\\left\{{}\begin{matrix}5-x< 0\\x+8< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5>x\ge-8\left(nhận\right)\\-8>x>5\left(VLý\right)\end{matrix}\right.\)
A = - (x + 18/1203)2 - 183/121
-(x+ 18/1203)2 ≤ 0
⇔ A(max) = -183/121 ⇔ x = -18/1203 = - 6/401
B = \(\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\)
để B(max) thì (x+1/3)2 + 5 nhỏ nhất
đặt C = (x+1/3)2 + 5 thì B(max) ⇔ C(min)
vì (x+1/3)2 ≥ 0 ⇔ C(min) = 5 ⇔ x =-1/3
B(max) = 4/5 ⇔ x = -1/3
a) \(A=-\left(\dfrac{x+18}{1203}\right)^2-\dfrac{183}{121}\le-\dfrac{183}{121}\)
Dấu bằng xảy ra
\(\Leftrightarrow x+18=0\Leftrightarrow x=-18\)
b) \(B=\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\)
Ta có : \(\left(x+\dfrac{1}{3}\right)^2+5\ge5\)
\(\Leftrightarrow\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\le\dfrac{4}{5}\)
Dấu bằng xảy ra
\(\Leftrightarrow x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\)