\(\dfrac{\text{9n+15}}{\text{3n-2}}\)
a,\(\dfrac{x}{-3}\text{=}\dfrac{-5}{15}\) b,\(\dfrac{1173}{x}\text{=}\dfrac{3}{5}\)
c,\(\dfrac{2}{x}\text{=}\dfrac{y}{15}\text{=}\dfrac{-25}{75}\)
a) \(x=\dfrac{-3.\left(-5\right)}{15}=1\)
b) \(x=\dfrac{5.1173}{3}=1955\)
c) Tách làm 2 phép tính
(1) \(\dfrac{y}{15}=\dfrac{-25}{75}\rightarrow y=\dfrac{-25.15}{75}=-5\)
(2) \(\dfrac{2}{x}=\dfrac{-25}{75}\rightarrow x=\dfrac{75.2}{-25}=-6\)
tìm số nguyên n để 9n+3/3n+1 thuộc Z
\(\frac{9n+3}{3n+1}=\frac{3\left(3n+1\right)}{3n+1}=3\)
Vậy với \(n\in Z\) thì \(\frac{9n+3}{3n+1}\in Z\)
Tính các giới hạn sau:
a) \(\lim\limits\dfrac{5n^3-3n^2+1}{1-3n^3}\)
b) \(\lim\limits\dfrac{-9n+5}{3n-3}\)
`a)lim[5n^3-3n^2+1]/[1-3n^3]`
`=lim[5-3/n+1/[n^3]]/[1/[n^3]-3]`
`=5/[-3]=-5/3`
_____________________________
`b)lim[-9n+5]/[3n-3]`
`=lim[-9+5/n]/[3-3/n]`
`=[-9]/3=-3`
Tìm \(n\text{∈}Z\) để \(A=\dfrac{3n+2}{n-1}\text{∈}Z\)
Ta có:
\(A=\dfrac{3n+2}{n-1}=\dfrac{\left(3n-3\right)+5}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{5}{n-1}=\dfrac{3\left(n-1\right)}{n-1}+\dfrac{5}{n-1}=3+\dfrac{5}{n-1}\)
Để \(A\in Z\Rightarrow\dfrac{5}{n-1}\in Z\Rightarrow5⋮n-1\) hay \(n-1\in U\left(5\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng giá trị:
\(n-1\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
\(n\) | \(2\) | \(0\) | \(6\) | \(-4\) |
Vậy với \(n\in\left\{-4;0;2;6\right\}\) thì \(\dfrac{3n+2}{n-1}\in Z\)
Để \(A\in Z\) thì \(3n+2⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5\) \(⋮n-1\)
Vì \(3\left(n-1\right)⋮n-1\)
\(\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)\)
mà \(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
\(n-1\) | 1 | \(-1\) | 5 | \(-5\) |
\(n\) | 2 | 0 | 6 | \(-4\) |
Kết luận | nhận | nhận | nhận | nhận |
Vậy \(n\in\left\{-4;0;2;6\right\}\).
Để \(A\in Z\Rightarrow3n+2⋮n-1\)
Ta có: \(\left\{{}\begin{matrix}3n+1⋮n-1\\n-1⋮n-1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3n+1⋮n-1\\3n-3⋮n-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3n-3+4⋮n-1\\3n-3⋮n-1\end{matrix}\right.\)
\(\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)\)
\(\Rightarrow n-1\in\left\{\pm1;\pm2;\pm3\right\}\)
\(\Rightarrow n\in\left\{-2;-1;0;2;3;4;\right\}\)
Giải phương trình: \(\dfrac{1}{\text{x^2-3x+2}}+\dfrac{1}{\text{x^2-5x+6}}+\dfrac{1}{\text{x^2-7x+12}}+\dfrac{1}{\text{x^2-9x+20}}=\dfrac{1}{15}\) có tập nghiệm là:
A. S={-1;5} B. S={11} C. S=\(\varnothing\) D. S={11;-5}
\(đkxđ:x\ne1;2;3;4;5\\ \Leftrightarrow\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}=\dfrac{1}{15}\\ \Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}=\dfrac{1}{15}\\ \Leftrightarrow\dfrac{1}{x-5}-\dfrac{1}{x-1}=\dfrac{1}{15}\\ \Leftrightarrow60=x^2-6x+5\\ \)
\(\Leftrightarrow60=x^2-6x+5\\ \Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\\ \Rightarrow D\)
\(\dfrac{121m+9n}{-10m-3n}\) với \(\dfrac{m}{3}\)=\(\dfrac{n}{5}\)
Đặt \(\dfrac{m}{3}=\dfrac{n}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=3k\\n=5k\end{matrix}\right.\)
Ta có: \(\dfrac{121m+9n}{-10m-3n}\)
\(=\dfrac{121\cdot3k+9\cdot5k}{-10\cdot3k-3\cdot5k}=\dfrac{363k+45k}{-30k-15k}\)
\(=\dfrac{408k}{-45k}=-\dfrac{136}{15}\)
BT1: Tìm x, biết:
5) \(\text{|}x+\dfrac{1}{3}\text{|}+\text{|}x+\dfrac{1}{5}\text{|}+\text{|}x+\dfrac{1}{15}\text{|}=4x\)
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{5}\right|\ge0\\\left|x+\dfrac{1}{15}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{5}+x+\dfrac{1}{15}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy ..
BT3: Tìm x, biết
15) \(2\text{|}\dfrac{1}{2}x-\dfrac{1}{3}\text{|}-\dfrac{3}{2}=\dfrac{1}{4}\)
16) \(\dfrac{3}{4}-2.\text{|}2x-\dfrac{2}{3}\text{|}=2\)
(Lưu ý: Dấu '' | '' là dấu giá trị tuyệt đối)
1) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{7}{8}+\dfrac{1}{3}\\\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{13}{24}\\\dfrac{1}{2}x=\dfrac{29}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-\dfrac{13}{24}\right):\dfrac{1}{2}\\x=\dfrac{29}{24}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{12}\\x=\dfrac{29}{12}\end{matrix}\right.\)
2) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)
\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)
\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)
\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}:2\)
\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{-5}{16}\\2x-\dfrac{2}{3}=\dfrac{5}{16}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-5}{16}+\dfrac{2}{3}\\2x=\dfrac{5}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{48}\\2x=\dfrac{47}{48}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{48}:2\\x=\dfrac{47}{48}:2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{96}\\x=\dfrac{47}{96}\end{matrix}\right.\)
Cho : \(\text{A}=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8\cdot3^8}{2^{15}}\)
\(\text{B}=\dfrac{6^2}{2^{16}}\)
Tính \(\dfrac{\text{A}}{\text{B}}\).
\(A=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8.3^8}{2^{15}}=\dfrac{3}{2^{15}}\left(1\right)\)
\(B=\dfrac{6^2}{2^{16}}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{3}{2^{15}}}{\dfrac{6^2}{2^{16}}}=\dfrac{1}{6}\)