a) \(\left|x-7\right|\ge x-7\Rightarrow A\ge x-7+3-x=-4\)

Dấu "=" xảy ra <=> \(x-7\ge0\Leftrightarrow x\ge7\)

b)\(\left|x+7\right|\ge x+7;\left|x+3\right|\ge0;\left|x+1\right|\ge-x-1\Rightarrow B\ge x+7+0-x-1=6\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+7\ge0\\x+3=0\\x+1\le0\end{cases}\Leftrightarrow x=-3}\)

c) \(\left|2-x\right|\ge x-2;\left|5-x\right|\ge5-x\Rightarrow C\ge x-2+5-x=3\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}2-x\le0\\5-x\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge2\\x\le5\end{cases}}\)

Lời giải:

$A=|x-2|+|y+3|=|2+y-2|+|y+3|=|y|+|y+3|$

$=|-y|+|y+3|\geq |-y+y+3|=3$

Vậy $A_{\min}=3$

Giá trị này đạt được khi $(-y)(y+3)\geq 0$

$\Leftrightarrow -3\leq y\leq 0$

=100000000007

100000000007

t nha

trả lời

là bằng 2

uy tín đến mức chưa có 10 sp  kaka

2+1.0=2

a: \(\left(-1\right)\left(-1\right)^2\cdot\left(-1\right)^3\cdot\ldots\cdot\left(-1\right)^{2014}\)

\(=\left(-1\right)^{1+2+\cdots+2014}\)

\(=\left(-1\right)^{2014\cdot\frac{2015}{2}}=\left(-1\right)^{1007\cdot2015}=-1\)

b: \(\frac17\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{130}+\frac{13}{60}\right)\)

\(=\frac17\left(\frac52+\frac{4}{11}+\frac{3}{22}+\frac{11}{130}+\frac{13}{60}\right)\)

\(=\frac17\left(3+\frac{13}{60}+\frac{11}{130}\right)=\frac17\cdot\frac{515}{156}=\frac{515}{1092}\)

Bài 1 :

\(\frac{11}{5}+\frac{22}{10}=\frac{22}{10}+\frac{22}{10}=\frac{44}{10}=\frac{22}{5}\)

\(\frac{66}{30}-\frac{70}{15}=\frac{66}{30}-\frac{140}{30}=\frac{-74}{30}=\frac{-37}{15}\)

Bài 2:

Ta có: \(\frac{11}{12}< 1< \frac{22}{3}\)

\(\Rightarrow\frac{11}{12}< \frac{22}{3}\)

Ta có: \(\frac{100}{22}< \frac{100}{20}=5< \frac{66}{3}=22\)

\(\Rightarrow\frac{100}{22}< \frac{66}{3}\)

Bài 3:

\(\frac{1}{2}\notinℤ\)

\(2,5\inℚ\)

\(\frac{13}{3}\notinℤ\)

\(ℕ\subsetℤ\subsetℚ\)