0,13/x =9/42
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ta có
\(\left|x+1\right|\ge0\Leftrightarrow\left|x+1\right|+3\ge3\Rightarrow GTNN\text{ }A=3\)
dấu bằng xảy ra khi \(\left|x+1\right|=0\Leftrightarrow x=-1\)
\(\left|x-\frac{2}{3}\right|\ge0\Leftrightarrow2\left|x-\frac{2}{3}\right|-1\ge-1\Rightarrow GTNN\text{ }B=-1\)
dấu bằng xảy ra khi \(\left|x-\frac{2}{3}\right|=0\Leftrightarrow x=\frac{2}{3}\)
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{n!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{n-1!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
\(=\frac{n!-1}{n!}\)
\(\frac{3}{7}\cdot\frac{11}{13}-\frac{3}{13}\cdot\frac{15}{7}-\frac{17}{7}=\frac{33}{91}-\frac{45}{91}-\frac{221}{91}=\frac{33-45-221}{91}=-\frac{233}{91}\)
a) \(=\frac{25}{36}\)
b) \(=\frac{50}{7}\)
c) \(=\frac{200}{279}\)
a) 1,2: 3,24 = 120 : 324 = 10:27
b) 215:34215:34 = 115:34=115.43=44:15115:34=115.43=44:15
c) 27:0,4227:0,42 = 27:42100=27.10042=200294=100147=100:14727:42100=27.10042=200294=100147=100:147
\(\frac{0,13}{x}=\frac{9}{42}\)
=> \(\frac{0,13}{x}=\frac{3}{14}\)
=> \(0,13\cdot14=3x\)
=> \(3x=1,82\)
=> \(x\approx0,6\)