\(\frac{x}{3}=\frac{y}{-5}=\frac{z}{2};4x-3z=-18\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\Rightarrow\) \(\frac{4x}{12}=\frac{3z}{6}=\frac{4x-3z}{12-6}=\frac{-18}{6}=-3\)

suy ra \(\hept{\begin{cases}\frac{x}{3}=-3\Leftrightarrow x=-9\\\frac{y}{-5}=-3\Leftrightarrow y=15\\\frac{z}{2}=-3\Leftrightarrow z=-6\end{cases}}\)

Vậy, ...

\(\frac{x}{3}=\frac{y}{-5}=\frac{z}{2}\) và 4x - 3z = -18

\(\Rightarrow\frac{4x}{12}=\frac{y}{-5}=\frac{3z}{6}=\frac{4x-3z}{12-6}=\frac{-18}{6}=-3\)

\(\rightarrow\hept{\begin{cases}\frac{x}{3}=-3\Rightarrow x=-9\\\frac{y}{-5}=-3\Rightarrow y=15\\\frac{z}{2}=-3\Rightarrow z=6\end{cases}}\)

gọi x,y ,z là số tờ bạc 500 đồng 2000 đ, 5000 đ.ta có x+y+z=540 và 500x=2000y=5000z.từ đó x=400,y=100,z=40

gọi a b c ll là số tờ tiền của 500d 2000d 5000d

ta co 500a=2000b=5000cvà a+b+c=54

=>a/10=b/2,5=c/1=54/13,5=4

=>a=40

b=10

c=4

1 Tính : 

a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)

\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)

\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)

\(=\frac{1}{n}\)

b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)

\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)

\(=\frac{3}{5}+\frac{1}{n}\)

c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)

\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Rightarrow C=1-B\left(1\right)\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

Lấy 2B trừ B ta có : 

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(B=1-\frac{1}{2^{10}}\left(2\right)\)

Thay (2) vào (1) ta có :

\(C=1-\left(1-\frac{1}{10}\right)\)

\(=1-1+\frac{1}{10}\)

\(=\frac{1}{10}\)

Vậy \(C=\frac{1}{10}\)

                                                             Bài giải

a, \(A=8-\left|x-1\right|\) đạt GTNN khi \(\left|x-1\right|\)đạt GTNN

Mà \(\left|x-1\right|\ge0\) Dấu " = " xảy ra khi \(\left|x-1\right|=0\) \(\Rightarrow\text{ }x-1=0\) \(\Rightarrow\text{ }x=1\)

\(\Rightarrow\text{ }A=8-\left|x-1\right|\le8\)

Vậy \(Max\text{ }A=8\)

                                                             Bài giải

a, \(A=8-\left|x-1\right|\) đạt GTNN khi \(\left|x-1\right|\)đạt GTNN

Mà \(\left|x-1\right|\ge0\) Dấu " = " xảy ra khi \(\left|x-1\right|=0\) \(\Rightarrow\text{ }x-1=0\) \(\Rightarrow\text{ }x=1\)

\(\Rightarrow\text{ }A=8-\left|x-1\right|\le8\)

Vậy \(Max\text{ }A=8\)

2 số -2 và -2

a, Tự chép đề bài :v

=> 2^2x-3 = ( 8³. 16⁵ ) : 4¹⁰

2^2x-3 = ( 2⁹. 2²⁰ ) : 2²⁰

2^2x-3 = 2²⁹ : 2²⁰

2^2x-3 = 2⁹

=> 2x - 3 = 9

2x = 9 + 3

2x = 12

x = 6 

Vậy....

b, 7. 2^x = 2⁹ + 5. 2⁸

7. 2^x = 1792

2^x = 1792 : 7

2^x = 256

2^x = 2⁸

=> x = 8 

Vậy ....

a  \(\frac{2^{2x-3}}{4^{10}}=8^3.16^5\)

\(\Leftrightarrow\frac{2^{2x-3}}{4^{10}}=2^{29}\)

\(\Leftrightarrow2^{2x-3}=2^{29}.4^{10}\)

\(\Leftrightarrow2^{2x-3}=2^{49}\)

\(\Leftrightarrow2x-3=49\)

\(\Leftrightarrow x=26\)

b  \(7.2^x=2^9+5.2^8\)

\(\Leftrightarrow7.2^x=2^8.(2+5)\)

\(\Leftrightarrow7.2^x=7.2^8\)

\(\Leftrightarrow x=7\)