\(\frac{5}{x}-\frac{y}{3}=\frac{1}{6}\)

\(\Leftrightarrow\frac{5.3}{3x}-\frac{y}{3}=\frac{1}{6}\)

\(\Leftrightarrow\frac{5.3}{3x}=\frac{1}{6}+\frac{y}{3}=\frac{1}{6}+\frac{2y}{6}\)

\(\Leftrightarrow\frac{15}{3x}=\frac{1+2y}{6}\)

\(\Rightarrow\hept{\begin{cases}15=1+2y\\3x=6\end{cases}\Rightarrow\hept{\begin{cases}15=1+2y\\x=2\end{cases}}}\)

Thế x = 2 vào,ta có: 

\(\frac{15}{3.2}=\frac{15}{6}=\frac{1.2y}{6}\)

\(\Leftrightarrow\frac{15}{6}=\frac{2y}{6}\Rightarrow y=15:2=7,5=8\)

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

7/x - y/1

=> xy = 7

=> x;y thuộc Ư(7) mà x;y nguyên 

=> x;y thuộc {1; 7; -1; -7}

xét bảng

tại sao \(\frac{7}{y}-\frac{y}{1}\)

\(\frac{x+1}{2}-\frac{3}{5}=\frac{1}{2y}\Leftrightarrow\frac{5x-1}{10}=\frac{1}{2y}\)

\(\Leftrightarrow2y\left(5x-1\right)=10\Leftrightarrow y\left(5x-1\right)=5\)

TH1: \(\left\{{}\begin{matrix}y=5\\5x-1=1\end{matrix}\right.\) \(\Rightarrow x=\frac{2}{5}\left(l\right)\)

TH2: \(\left\{{}\begin{matrix}y=1\\5x-1=5\end{matrix}\right.\) \(\Rightarrow x=\frac{6}{5}\left(l\right)\)

TH3: \(\left\{{}\begin{matrix}y=-1\\5x-1=-5\end{matrix}\right.\) \(\Rightarrow x=-\frac{4}{5}\left(l\right)\)

TH4: \(\left\{{}\begin{matrix}y=-5\\5x-1=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;-5\right)\)

Lên đây liếc ngó mấy bài toán để 9/6 thi chuyên :''<

Đại ca dạy em học Toán với, tỉ lệ 1 chọi 5 em lo quá :((

khó @gmail.com

nè, không làm thôi ằ nhagg. khó thì đừng gửi câu trả lời làm gì cho mệt nha bạn 

\(\frac{x}{5}-\frac{1}{y+2}=\frac{1}{10}\)

\(\frac{1}{y+2}=\frac{x}{5}-\frac{1}{10}=\frac{2x}{10}-\frac{1}{10}=\frac{2x-1}{10}\)

\(\Rightarrow\left(y+2\right).\left(2x-1\right)=1.10=10\)

\(\Rightarrow2x-1\inƯ\left(10\right)\)

Mà 2x - 1 là lẻ

\(\Rightarrow2x-1\in\left[1;5;-1;-5\right]\)

Xét \(2x-1=1\Rightarrow x=1\) 

\(\Rightarrow y+2=10\Rightarrow y=8\)

Xét \(2x-1=5\Rightarrow x=3\)

\(\Rightarrow y+2=2\Rightarrow y=0\)

Xét \(2x-1=-1\Rightarrow x=0\)

\(\Rightarrow y+2=-10\Rightarrow y=-12\)

Xét \(2x-1=-5\Rightarrow x=-2\)

\(\Rightarrow y+2=-2\Rightarrow y=-4\)

tính: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1982}+\frac{1}{1984}+\frac{1}{1986}\)

Bài 2 

\(a,\)\(\left(x^2+7\right)\left(x^2-49\right)< 0\)

Vì \(x^2+7>0\)\(\Rightarrow x^2-49< 0\)

\(\Rightarrow\left(x-7\right)\left(x+7\right)< 0\)

\(...\)

Bài 2:

a) \(\left(x^2+7\right).\left(x^2-49\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x^2+7< 0\\x^2-49>0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+7>0\\x^2-49< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2< -7\\x^2>49\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}x^2>-7\\x^2< 49\end{cases}}\)

\(\Leftrightarrow-7< x^2< 49\)

Mà \(x^2\ge0\)và  \(x^2\)là 1 SCP

\(\Rightarrow x^2\in\left\{1;4;9;16;25;36\right\}\)

\(\Rightarrow x\in\left\{1;2;3;4;5;6\right\}\)

Vậy \(x\in\left\{1;2;3;4;5;6\right\}\)