Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu a:
\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)
-8.(-7) = 4.(3\(x\) - 1)
56 = 12\(x\) - 4
12\(x\) = 56+ 4
12\(x\) = 60
\(x\) = 60 : 12
\(x\) = 5
Vậy \(x\) = 5
Câu b:
\(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(x^2\) = (-3)\(^2\)
\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)
Vậy \(x\in\left\lbrace-3;3\right\rbrace\)
Câu c:
\(-\frac{4}{y}=\frac{x}{2}\)
-4.2 = \(x.y\)
\(xy=-8\)
Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}
Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)
Câu 2:
(\(x-1)\)(y + 2) = 7
Ư(7) = {-7; -1; 1; 7}
Lập bảng ta có:
\(x\)-1 | -7 | -1 | 1 | 7 |
\(x\) | -6 | 0 | 2 | 8 |
y+2 | -1 | -7 | 7 | 1 |
y | -3 | -9 | 5 | -1 |
\(x;y\in Z\) | tm | tm | tm | tm |
Theo bảng trên ta có:
(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)
Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)
1. Ta có:\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+2b-3c}{2+6-12}=\frac{-20}{-4}=5\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{2}=5\\\frac{b}{3}=5\\\frac{c}{4}=5\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=10\\b=15\\c=20\end{cases}}\)
2. Ta có:\(\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15}\)
\(\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}\)
\(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a-b+c}{10-15+12}=\frac{-49}{7}=-7\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{10}=-7\\\frac{b}{15}=-7\\\frac{c}{12}=-7\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=-70\\b=-105\\c=-84\end{cases}}\)
1. Ta có:a2 =b3 =c4 =a+2b−3c2+6−12 =−20−4 =5
| a2 =5 |
| b3 =5 |
| c4 =5 |
| a=10 |
| b=15 |
| c=20 |
2. Ta có:a2 =b3 ⇒a10 =b15
b5 =c4 ⇒b15 =c12
⇒a10 =b15 =c12 =a−b+c10−15+12 =−497 =−7
| a10 =−7 |
| b15 =−7 |
| c12 =−7 |
| a=−70 |
| b=−105 |
| c=−84 |
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
Bài 1 :
Theo bài ra ta có : \(\frac{a}{b}=\frac{2}{3}\Leftrightarrow\frac{a}{2}=\frac{b}{3}\)
Áp dụng t/c dãy tỉ số ''='' nhau ta có
\(\frac{a}{2}=\frac{b}{3}=\frac{a+b}{2+3}=\frac{10}{5}=2\)
\(\Leftrightarrow\frac{a}{2}=2\Leftrightarrow a=4\)
\(\Leftrightarrow\frac{b}{3}=2\Leftrightarrow b=6\)
Bài 2 :
Tìm khó quá cj thử x2;x3 ko ra rồi )):
\(1.\frac{4}{3}.a=\frac{1}{50}\)
\(a=\frac{1}{50}:\frac{4}{3}\)
\(a=\frac{3}{200}\)
\(2.\frac{1}{20}.a=4\)
\(a=4:\frac{1}{20}\)
\(a=80\)
bài 1
đổi 2%=1^50
ta có :4^3.a=1^50
=>a=1^50:4^3
=>a=1^50.3^4
=>a=3^200
bài 2
đổi 5%=1^20
ta có :1^20.a=4
=>a= 4.20
=>a=80
\(\frac{1}{50}\)chứ ko phải 150 nha bn ^ là mũ