\(\frac{x}{10}=\frac{y}{7}=\frac{z}{23}\) và \(2x+y-z=12\)

Đặt : \(\frac{x}{10}=\frac{y}{7}=\frac{z}{23}=k\)

\(\Rightarrow\hept{\begin{cases}x=10k\left(1\right)\\y=7k\left(2\right)\\z=23k\left(3\right)\end{cases}}\)

Lại có : \(2x+y-z=12\left(4\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\)và \(\left(4\right)\)suy ra :

\(2.10k+7k-23k=12\)

\(20k+7k-23k=12\)

\(4k=12\)

\(k=3\)

Thay \(k=3\)vào \(\left(1\right),\left(2\right),\left(3\right)\)ta được :

\(\Rightarrow\hept{\begin{cases}x=10.3=30\\y=7.3=21\\z=23.3=69\end{cases}}\)

Vậy ...................

\(\frac{0,13}{x}=\frac{9}{42}\)

=>   \(\frac{0,13}{x}=\frac{3}{14}\)

=>   \(0,13\cdot14=3x\)

=>  \(3x=1,82\)

=>  \(x\approx0,6\)

mình chưa làm tập hai xin lỗi nhé 

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