xin lỗi\(^{100}\)

\(B=\left|x+10\right|+\left|y-10\right|+178\) ( \(x,y\) ∈ Z )

Ta có :

\(\left|x+10\right|\ge0\) ∀ \(x\)

\(\left|y-10\right|\ge0\) ∀ \(y\)

\(\Rightarrow\left|x+10\right|+\left|y-10\right|\ge0\)

\(\Rightarrow\left|x+10\right|+\left|y-10\right|+178\ge0+178\)

\(\Rightarrow B\ge178\) . Dấu '' = '' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+10=0\\y-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0-10=-10\left(TM\right)\\y=0+10=10\left(TM\right)\end{matrix}\right.\)

Vậy Min B = 178 \(\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=10\end{matrix}\right.\)

2.

A = 5 - \(\left|x-3\right|\)

Ta có :

\(-\left|x-3\right|\le0\) ∀ \(x\)

\(\Rightarrow5-\left|x-3\right|\le0+5\)

\(\Rightarrow A\le5\) . Dấu '' = '' xảy ra ⇔ x - 3 =0

⇔ x = 0 + 3 = 3

Vậy Max A = 5 ⇔ x = 3

Bài 4:

\(A=2x^2-15\ge-15\\ A_{min}=-15\Leftrightarrow x=0\\ B=2\left(x+1\right)^2-17\ge-17\\ B_{min}=-17\Leftrightarrow x=-1\)

Bài 5:

\(A=-x^2+14\le14\\ A_{max}=14\Leftrightarrow x=0\\ B=25-\left(x-2\right)^2\le25\\ B_{max}=25\Leftrightarrow x=2\)

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

Ai giúp mik vs

Huhu ai giúp vs

\(A=5-8x+x^2=-8x+x^2+6-11\)

\(=\left(x-4\right)^2-11\)

Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy Amin = - 11 <=> x = 4

\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)

\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy Bmax = 9 <=> x = - 1

a) \(M=2022-\left|x-9\right|\le2022\)

\(maxM=2022\Leftrightarrow x=9\)

b) \(N=\left|x-2021\right|+2022\ge2022\)

\(minN=2022\Leftrightarrow x=2021\)

2: B=|x+5|-|x-2|<=|x+5-x+2|=7

Dấu = xảy ra khi -5<=x<=2