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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+2}{2017}+\frac{x+3}{2016}+\frac{x+4}{2015}+\frac{x+5}{1007}+\frac{x+2074}{11}=0\)
\(\Leftrightarrow\frac{x+2}{2017}+1+\frac{x+3}{2016}+1+\frac{x+4}{2015}+1+\frac{x+5}{1007}+2+\frac{x+2074}{11}-5=0\)
\(\Leftrightarrow\frac{x+2019}{2017}+\frac{x+2019}{2016}+\frac{x+2019}{2015}+\frac{x+2019}{1007}+\frac{x+2019}{11}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{1007}+\frac{1}{11}\right)=0\)
\(\Leftrightarrow\left(x+2019\right)=0vì\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{1007}+\frac{1}{11}\right)\ne0\)
\(\Leftrightarrow x=-2019\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3x-2}{8}=\frac{5y+6}{3}=\frac{3x-5y-8}{8-3}=\frac{3x-5y-8}{5}\)
\(+,3x=5y+8\Rightarrow\frac{5y+6}{8}=\frac{5y+6}{3}\Rightarrow y=-\frac{6}{5}\Rightarrow x=\frac{2}{3}\)
\(+,3x\ne5y+8\Rightarrow5=10x\Leftrightarrow x=\frac{1}{2}\Rightarrow\frac{-1}{16}=\frac{5y+6}{3}\Rightarrow....\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)Ta có: \(2^{x+2}+2^{x+5}-36864=0\) \(\Rightarrow\) \(2^{x+2}+2^{x+5}=36864\)
\(\Rightarrow\) \(2^x\times2^2+2^x\times2^5=36864\)
\(\Rightarrow\) \(2^x\times\left(2^2+2^5\right)=36864\)
\(\Rightarrow\) \(2^x\times\left(4+32\right)=36864\)
\(\Rightarrow\) \(2^x\times36=36864\)
\(\Rightarrow\) \(2^x=36864\div36=1024\)
Mà \(1024=2^{10}\) \(\Rightarrow\)\(x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Tìm các số nguyên x,y thỏa mãn
*\(2xy+6x-y=10\)
\(\Leftrightarrow\left(2xy+6x\right)-y-3=10-3=7\)
\(\Leftrightarrow2x\left(y+3\right)-\left(y+3\right)=7\)
\(\Leftrightarrow\left(y+3\right)\left(2x-1\right)=7\)
Lập bảng xét ước nữa là xong.
* \(xy+4x-3y=1\Leftrightarrow\left(xy+4x\right)-3y-12=1-12=-11\)
\(\Leftrightarrow x\left(y+4\right)-\left(3y+12\right)=-11\)
\(\Leftrightarrow x\left(y+4\right)-3\left(y+4\right)=-11\)
\(\Leftrightarrow\left(x-3\right)\left(y+4\right)=-11\)
Lập bảng xét ước nữa là xong.
Mới nhìn vào thấy bài toán hay hay lạ kì.
Thêm một vào bớt một ra
Tức thì bài toán trở nên dễ dàng:
\(\frac{x}{50}-\frac{x-1}{51}=\frac{x+2}{48}-\frac{x-3}{53}\)
\(\Leftrightarrow\frac{x}{50}+1-\frac{x-1}{51}-1=\frac{x+2}{48}+1-\frac{x-3}{53}-1\)
\(\Leftrightarrow\left(\frac{x}{50}+1\right)-\left(\frac{x-1}{51}+1\right)=\left(\frac{x+2}{48}+1\right)-\left(\frac{x-3}{53}+1\right)\)
\(\Leftrightarrow\frac{x+50}{50}-\frac{x+50}{51}=\frac{x+50}{48}-\frac{x+50}{53}\)
\(\Leftrightarrow\frac{x+50}{50}-\frac{x+50}{51}-\frac{x+50}{48}+\frac{x+50}{53}=0\)
\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{50}-\frac{1}{51}-\frac{1}{48}+\frac{1}{53}\right)=0\)
Dễ thấy \(\left(\frac{1}{50}-\frac{1}{51}-\frac{1}{48}+\frac{1}{53}\right)\ne0\)
Do đó x + 50 = 0 hay x = -50
![](https://rs.olm.vn/images/avt/0.png?1311)
3. Tìm x biết: |15-|4.x||=2019
\(\Rightarrow\orbr{\begin{cases}15-\left|4x\right|=2019\\15-\left|4x\right|=-2019\end{cases}\Rightarrow\orbr{\begin{cases}\left|4x\right|=-2004\\\left|4x\right|=2034\end{cases}}}\)
vì \(4x\ge0\)\(\Rightarrow\)|4x|=2043\(\Rightarrow4x=2034\Rightarrow x=508,5\)
KL: x=508,5
![](https://rs.olm.vn/images/avt/0.png?1311)
a/\(\frac{y}{5}+\frac{1}{10}=\frac{1}{x}\)
\(\frac{y.2}{10}+\frac{1}{10}=\frac{1}{x}\)
\(\frac{y.2+1}{10}=\frac{1}{x}\Leftrightarrow\left(y.2+1\right)x=10\)
Ta có Ư(10)={-1;1;-2;2-5;5-10;10}
Mà y.2+1 là số lẻ nên có bảng sau:
\(y.2+1\) | \(-1\) | \(1\) | \(-5\) | \(5\) |
\(y.2\) | \(-2\) | \(0\) | \(-6\) | \(4\) |
\(y\) | \(-1\) | \(0\) | \(-3\) | \(2\) |
\(x\) | \(-10\) | \(10\) | \(-2\) | \(2\) |
b/\(\frac{x}{4}-\frac{1}{2}=\frac{3}{y}\)
\(\frac{x}{4}-\frac{2}{4}=\frac{3}{y}\)
\(\frac{x-2}{4}=\frac{3}{y}\Leftrightarrow\left(x-2\right)y=12\)
Ta có Ư(12)={-1;1;-2;2-3;3;-4;4;-6;6;-12;12}
Ta có bảng sau:
x-2 | -1 | 1 | -2 | 2 | -3 | 3 | -4 | 4 | -6 | 6 | -12 | 12 |
x | 1 | 3 | 0 | 4 | -1 | 5 | -2 | 6 | -4 | 8 | -10 | 14 |
y | -12 | 12 | -6 | 6 | -4 | 4 | -3 | 3 | -2 | 2 | -1 | 1 |
CHÚC BẠN HỌC TỐT!!!