Question

giải hộ tui

Answer 1

a) Ta có: \(\left|x+1\right|\ge0\)

\(\Rightarrow A=\left|x+1\right|+5\ge5\)

Dấu " = " khi \(x+1=0\Rightarrow x=-1\)

Vậy \(MIN_A=5\) khi x = -1

b) Ta có: \(B=\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B\ge\left|x-1+3-x\right|=\left|-2\right|=2\)

Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\Rightarrow1\le x\le3\)

Vậy \(MIN_B=2\) khi \(1\le x\le3\)

c) Ta có: \(C=x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5\)

\(\left(x+1\right)^2\ge0\)

\(\Rightarrow C=\left(x+1\right)^2+5\ge5\)

Dấu " = " khi \(x+1=0\Rightarrow x=-1\)

Vậy \(MIN_C=5\) khi x = -1

d) \(D=x^2-2x+7=x^2-2x+1+6=\left(x-1\right)^2+6\)

Ta có: \(\left(x-1\right)^2\ge0\)

\(\Rightarrow D=\left(x-1\right)^2+6\ge6\)

Dấu " = " khi \(x-1=0\Rightarrow x=1\)

Vậy \(MIN_B=6\) khi x = 1