1) \(-x-3=-2\left(x+7\right)\\ \Rightarrow-x-3=-2x-14\\ \Rightarrow-x+2x=-14+3\\ \Rightarrow x=-11\)

2) \(A=\frac{12}{\left(x+1\right)^2+3}\\ Tac\text{ó}:\left(x+1\right)^2\ge0\\ \Rightarrow\left(x+1\right)^2+3\ge3\\ \Rightarrow A\le\frac{12}{3}=4\)

Max A=4 khi x=-1

3) Đăt : \(n^2+4=k^2\\ \Rightarrow k^2-n^2=4\\ \Rightarrow\left(k-n\right)\left(k+n\right)=4\)

lập bang ra rồi tính

\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)

\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)

\(=\frac{4}{3}.\frac{1}{3}+4\)

\(=4+4\)

\(=8\)

1, Vì (x-2)² > 0

=> (x-1)² + 2008 > 2008

Dấu "=" xảy ra

<=> (x-1)² = 0

<=> x-1 = 0

<=> x = 1

KL: A_min = 2008 <=> x = 1

2, Vì |x+5| > 0

=> |x+5| + 2009 > 2009

Dấu "=" xảy ra

<=> |x+5| = 0

<=> x+5 = 0

<=> x = -5

KL: B_min = 2009 <=> x = -5

1, A = ( x - 1 )² + 2008

   Mà : ( x - 1 )² \(\ge\) 0

Mà : A_min => ( x - 1 )²

<=> x = 1

=> A_min  = 2008 khi x = 1

2, B = | x + 5 | + 2009

Vì : | x + 5 | \(\ge\) 0

Mà : B_min => | x + 5 | = 0

<=> x  = -5

=>  B_min = 2009 khi x = -5

1 . \(\frac{a}{-b}\) = \(\frac{-a}{b}\) vì a.b = -a.(-b)

2. tương tự