giúp mik nhoa mik đag cần gấp hihi

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) \(\dfrac{2}{x}=\dfrac{-25}{75}\rightarrow x=\dfrac{75.2}{-25}=-6\)

\(\frac{9n+3}{3n+1}=\frac{3\left(3n+1\right)}{3n+1}=3\)

Vậy với \(n\in Z\) thì  \(\frac{9n+3}{3n+1}\in Z\)

`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`

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`b)lim[-9n+5]/[3n-3]`

`=lim[-9+5/n]/[3-3/n]`

`=[-9]/3=-3`

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:

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:

Vậy \(n\in\left\{-4;0;2;6\right\}\).

Để \(A\in Z\Rightarrow3n+2⋮n-1\)

Ta có: \(\left\{{}\begin{matrix}3n+1⋮n-1\\n-1⋮n-1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3n+1⋮n-1\\3n-3⋮n-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3n-3+4⋮n-1\\3n-3⋮n-1\end{matrix}\right.\)

\(\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)\)

\(\Rightarrow n-1\in\left\{\pm1;\pm2;\pm3\right\}\)

\(\Rightarrow n\in\left\{-2;-1;0;2;3;4;\right\}\)

 \(đkxđ:x\ne1;2;3;4;5\\ \Leftrightarrow\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}=\dfrac{1}{15}\\ \Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}=\dfrac{1}{15}\\ \Leftrightarrow\dfrac{1}{x-5}-\dfrac{1}{x-1}=\dfrac{1}{15}\\ \Leftrightarrow60=x^2-6x+5\\ \) 

\(\Leftrightarrow60=x^2-6x+5\\ \Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\\ \Rightarrow D\)

câu a

với=với

Đặt \(\dfrac{m}{3}=\dfrac{n}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=3k\\n=5k\end{matrix}\right.\)

Ta có: \(\dfrac{121m+9n}{-10m-3n}\)

\(=\dfrac{121\cdot3k+9\cdot5k}{-10\cdot3k-3\cdot5k}=\dfrac{363k+45k}{-30k-15k}\)

\(=\dfrac{408k}{-45k}=-\dfrac{136}{15}\)

\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)

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 ..

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) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}:2\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{-5}{16}\\2x-\dfrac{2}{3}=\dfrac{5}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-5}{16}+\dfrac{2}{3}\\2x=\dfrac{5}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{48}\\2x=\dfrac{47}{48}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{48}:2\\x=\dfrac{47}{48}:2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{96}\\x=\dfrac{47}{96}\end{matrix}\right.\)

\(A=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8.3^8}{2^{15}}=\dfrac{3}{2^{15}}\left(1\right)\)

\(B=\dfrac{6^2}{2^{16}}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{3}{2^{15}}}{\dfrac{6^2}{2^{16}}}=\dfrac{1}{6}\)