Đặt \(\dfrac{m}{3}=\dfrac{n}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=3k\\n=5k\end{matrix}\right.\)
Ta có: \(\dfrac{121m+9n}{-10m-3n}\)
\(=\dfrac{121\cdot3k+9\cdot5k}{-10\cdot3k-3\cdot5k}=\dfrac{363k+45k}{-30k-15k}\)
\(=\dfrac{408k}{-45k}=-\dfrac{136}{15}\)
Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Cho A = \(\dfrac{3n+1}{2n+3}\) . Tìm n ∈ Z để A là số nguyên
\(\dfrac{help}{me}\)
Với a≠0, cho dãy số\(\dfrac{-2}{a};\dfrac{3}{a^3};\dfrac{-4}{a^5};\dfrac{5}{a^7};...\) biết số hạng thứ 5 của dãy số có dạng\(\dfrac{x}{a^5}\). \(x;n\)≠0. Giá trị của \(x+n=?\)
1,\(\dfrac{3}{16}\)- ( x - \(\dfrac{5}{4}\) ) - ( \(\dfrac{3}{4}\) + \(\dfrac{-7}{8}\) - 1 ) = \(2\dfrac{1}{2}\)
2,\(\dfrac{1}{2}\) . ( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\) ) = \(\dfrac{1}{5}\) - x + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\) )
Giúp mik nhanh với ạ .
Cho A=[\(\dfrac{n}{2}\)]+[\(\dfrac{n+1}{2}\)]
B=[\(\dfrac{n}{3}\)]+[\(\dfrac{n+1}{3}\)]+[\(\dfrac{n+2}{3}\)]
Với giá trị nào của nϵZ thì
A⋮2
B⋮3
1, Tính hợp lí
\(A=\dfrac{0.5+\dfrac{7}{12}-\dfrac{5}{6}}{1-\dfrac{2}{3}+0,75}\)
\(B=-66.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124.\left(-37\right)+126.\left(-62\right)\)
\(N=\left(60\dfrac{7}{13}+50\dfrac{8}{13}-11.\dfrac{2}{13}\right)x\) (Với \(x=-2017\dfrac{7}{10}\))
\(M=\left(1-\dfrac{2}{2.3}\right).\left(1-\dfrac{2}{3.4}\right).\left(1-\dfrac{2}{4.5}\right).....\left(1-\dfrac{2}{99.100}\right)\)
Tìm số nguyên n, biết rằng:
\(\dfrac{1}{4} . \dfrac{2}{6} . \dfrac{3}{8} .\dfrac{4}{10} . \dfrac{5}{12} .... \dfrac{30}{62} . \dfrac{31}{64} = 2^{n}\)\(\)
Tìm $x$, biết:
$a)$ $\dfrac{7}{4} x- \dfrac{3}{2}=- \dfrac{4}{5}$
$b)$ $\Big(x- \dfrac{1}{4}\Big)^2= \dfrac{5}{36}-\Big( \dfrac{1}{3}\Big)^2$
$c)$ $- x + \dfrac{3}{2}=x+\dfrac{3}{5}$
1. a) \(\dfrac{5}{3}+\dfrac{-2}{7}-\left(-1,2\right)\)
b) \(0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
Giúp mình với 
