Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\frac{7}{8}-\frac{1}{4}.\frac{5}{2}=\frac{x}{16}\)
\(\frac{7}{8}-\frac{5}{8}=\frac{x}{16}\)
\(\frac{2}{8}=\frac{x}{16}\)
\(\frac{4}{16}=\frac{x}{16}\)
=> X=4
k nha
\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)
\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)
\(\Rightarrow x\in\){9:10;11;12;13;14}
\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)
\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)
Vậy \(x\in\left\{11;12;13\right\}\)
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
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:
5n+7 chia hết cho 3n+2
=> [3(5n+7) - 5(3n + 2)] chia hết cho 3n+2
=> (15n + 21 - 15n - 10) chia hết cho 3n+2
=> 11 chia hết cho 3n + 2
=> 3n + 2 thuộc Ư(11) = {1;-1;11;-11}
Ta có bảng:
Vậy n = {-1;3}