a, \(\left|x\right|=-1,2\)

\(\Rightarrow x\in\varnothing\)

b, \(\left|x\right|+1,5=3,7\)

\(\left|x\right|=3,7-1,5\)

\(\left|x\right|=2,2\)

\(\Rightarrow x\in\left\{-2,2;2,2\right\}\)

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

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

\(\left|x+\frac{1}{3}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3-\frac{1}{3}\\x=-3-\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}\)

Câu 1 :

\(\left|x\right|=-1,2\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)

Câu 2 :

\(\left|x\right|+1,5=3,7\)

\(\left|x\right|=2,2\)

\(\Rightarrow x=2,2\) hoặc \(x=-2,2\)

Câu 3 :

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

\(\left|x+\frac{1}{3}\right|=3\)

TH 1:

\(x+\frac{1}{3}=3\)

\(x=\frac{8}{3}\)

TH 2:

\(x+\frac{1}{3}=-3\)

\(x=-\frac{10}{3}\)

Vậy \(x=\frac{8}{3}\) và \(x=-\frac{10}{3}\)

\(\left|x\right|=-1,2\)

Ta luôn có \(\left|x\right|\ge0\) \(\forall x.\)

\(\Rightarrow\left|x\right|>-1,2\)

\(\Rightarrow\left|x\right|\ne-1,2\)

Vậy \(x\in\varnothing.\)

b) \(\left|x\right|+1,5=3,7\)

\(\Rightarrow\left|x\right|=3,7-1,5\)

\(\Rightarrow\left|x\right|=2,2\)

\(\Rightarrow\left[{}\begin{matrix}x=2,2\\x=-2,2\end{matrix}\right.\)

Vậy \(x\in\left\{2,2;-2,2\right\}.\)

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

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

\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3-\frac{1}{3}\\x=\left(-3\right)-\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}.\)

Chúc bạn học tốt!

a) Vì \(\sqrt{x-5}\) ≥0

⇒ \(\sqrt{x-5}+7\) ≥ 7

Min A=7⇔x-5=0

             ⇔x=5

b) Vì \(\sqrt{3x-5}\) ≥0

⇒ 8-\(\sqrt{3x-5}\) ≤8

Max=8⇔3x-5\(=\)0

           ⇔\(x=\dfrac{5}{3}\)

\(A=\left|3,7-x\right|+2,5\ge2,5\)

\(MinA=2,5\Leftrightarrow3,7-x=0\Rightarrow x=3,7\)

\(\left|x+1,5\right|-4,5\ge-4,5\)

\(MinB=-4,5\Leftrightarrow x+1,5=0\Rightarrow x=-1,5\)

\(C=1,5-\left|x+1,1\right|\le1,5\)

\(MinC=1,5\Leftrightarrow x+1,1=0\Rightarrow x=-1,1\)

\(A=\left|3-x\right|+8\ge8\)

\(minA=8\Leftrightarrow x=3\)

\(B=\left|x+2\right|-4\ge-4\)

\(minB=-4\Leftrightarrow x=-2\)

\(A=\left|3-x\right|+8\ge8\forall x\)

Dấu '=' xảy ra khi x=3

\(B=\left|x+2\right|-4\ge-4\forall x\)

Dấu '=' xảy ra khi x=-2