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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

=-3/2 khi x=1/2

Đề là

\(C=\frac{3}{\left|x-1\right|+\left(x-1\right)4+1}+\frac{1}{2}.\)

hay là :

\(C=\frac{3}{\left|x-1\right|+\left(x-1\right)4+1+\frac{1}{2}}\)

\(C=\frac{3}{\left|x+1\right|+\left(x-1\right)^4+1}+\frac{1}{2}\)

Ta có: \(\hept{\begin{cases}\left|x-1\right|\ge0\forall x\\\left(x-1\right)^4\ge0\forall x\end{cases}}\)

\(\Rightarrow\left|x-1\right|+\left(x-1\right)^4\ge0\)

\(\Rightarrow\left|x-1\right|+\left(x-1\right)^4+1\ge1\)

\(\Rightarrow\frac{3}{\left|x-1\right|+\left(x-1\right)^4+1}\le\frac{3}{1}=3\)

\(\Rightarrow\frac{3}{\left|\text{x}-1\right|+\left(x-1\right)^4+1}+\frac{1}{2}\le3+\frac{1}{2}=\frac{7}{2}\)

hay \(MaxC=\frac{7}{2}\)

Dấu "=" xảy ra khi \(\left|x-1\right|=\left(x-1\right)^4=0\)

\(\Rightarrow x-1=0\)

\(x=1\)

Vậy \(MaxC=\frac{7}{2}\) tại \(x=1\).

câu hỏi tương tự nha kim hương tick mình nha bạn hiền