Vì | x - 3 | \(\ge\)0                                             ( 1 )

=> | x - 3 | + 2 \(\ge\)2

=> ( | x - 3 | + 2 )²  \(\ge\) 2² = 4

Vì | y + 3 | \(\ge\) 0                                          ( 2 )

Từ ( 1 ) và ( 2 ) => ( | x - 3 | + 2 )² + | y + 3 | + 2007 \(\ge\) 4 + 0 + 2007

                         => P \(\ge\) 2011

Dấu "=" xảy ra khi | x - 3 | = 0 và | y + 3 | = 0

                         => x - 3 = 0 và y + 3 = 0

                         => x = 3 và y = -3

Vậy GTNN của P là 2011 khi ( x ; y ) = ( 3 ; -3 )

lm dưới rùi

hi sư huynh đệ mới lớp 6

\(A=\frac{x^2-2x+2007}{2007x^2},\left(x\ne0\right)\)

\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}=\frac{x^2-2x.2007+2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}=\) \(\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)

\(A_{min}=\frac{2006}{2007}\) khi \(x-2007=0\) hay \(x=2007\)

Chúc bạn học tốt !!!

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12007" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">12007 - 22007x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">22007x + 1x2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">1x2

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<=> x = 2007

Vậy x = 2007 thì Amin

bài này từng có trên violimpic đấy bạn

n lớn nhất khi x=0

=>\(\left(1-2x\right)^2\ge0\)

=>\(\left(1-2x\right)^2+2007\ge2007\)

=>\(\frac{1}{\left(1-2x\right)^2+2007}\le\frac{1}{2007}\)

GTLN N = \(\frac{1}{2007}\)

dấu "=" xảy ra khi 1-2x=0 => x=\(\frac{1}{2}\)

vậy x=\(\frac{1}{2}\)thì GTLN N = \(\frac{1}{2007}\)

đợi tý

a) Để \(A=\dfrac{2022}{\left|x\right|+2023}\) đạt Max thì |x| + 2023 phải đạt Min

Ta có \(\left|x\right|\ge0\forall x\Rightarrow\left|x\right|+2023\ge2023\forall x\)

\(\Rightarrow\dfrac{2022}{\left|x\right|+2023}\le\dfrac{2022}{2023}\forall x\)

Dấu "=" xảy ra khi \(\left|x\right|=0\Rightarrow x=0\)

Vậy Max \(A=\dfrac{2022}{\left|x\right|+2023}=\dfrac{2022}{2023}\) đạt được khi x = 0

b) Để \(B=\left(\sqrt{x}+1\right)^{99}+2022\) đạt Min với \(x\ge0\) thì \(\sqrt{x}+1\) phải đạt Min

Ta có \(\sqrt{x}\ge0\forall x\ge0\Rightarrow\sqrt{x}+1\ge1\forall x\ge0\)

\(\Rightarrow\left(\sqrt{x}+1\right)^{99}+2022\ge1+2022\ge2023\forall x\ge0\)

Dấu "=" xảy ra khi \(\sqrt{x}=0\Rightarrow x=0\)

Vậy Max \(B=\left(\sqrt{x}+1\right)^{99}+2022=2023\) đạt được khi x = 0

Câu c) và d) thì tự làm, ko có rảnh =))))

Đã trả lời rồi còn độ tí đồ ngull

tach 14-x = 10-4-x roi sau do chac ban cung phai tu biet lam