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Câu a:
\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)
-8.(-7) = 4.(3\(x\) - 1)
56 = 12\(x\) - 4
12\(x\) = 56+ 4
12\(x\) = 60
\(x\) = 60 : 12
\(x\) = 5
Vậy \(x\) = 5
Câu b:
\(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(x^2\) = (-3)\(^2\)
\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)
Vậy \(x\in\left\lbrace-3;3\right\rbrace\)
Câu c:
\(-\frac{4}{y}=\frac{x}{2}\)
-4.2 = \(x.y\)
\(xy=-8\)
Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}
Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)
Câu 2:
(\(x-1)\)(y + 2) = 7
Ư(7) = {-7; -1; 1; 7}
Lập bảng ta có:
\(x\)-1 | -7 | -1 | 1 | 7 |
\(x\) | -6 | 0 | 2 | 8 |
y+2 | -1 | -7 | 7 | 1 |
y | -3 | -9 | 5 | -1 |
\(x;y\in Z\) | tm | tm | tm | tm |
Theo bảng trên ta có:
(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)
Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)
a) Ta có : \(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
\(\Rightarrow\frac{x}{3}-\frac{1}{5}=\frac{4}{y}\)
\(\Rightarrow\frac{x.5}{15}-\frac{3}{15}=\frac{4}{y}\)
\(\Rightarrow\frac{x.5-3}{15}=\frac{4}{y}\)
\(\Rightarrow\left(x.5-3\right).y=15.4\)
\(\Rightarrow x.5.y-3.5=60\)
\(\Rightarrow xy5-15=60\)
\(\Rightarrow xy5=60+15\)
\(\Rightarrow xy5=75\)
\(\Rightarrow xy=75\div5\)
\(\Rightarrow xy=15\)
\(\Rightarrow xy=1.15=3.5=\left(-15\right)\left(-1\right)=\left(-3\right)\left(-5\right)=\left(-5\right)\left(-3\right)=\left(-1\right)\left(-15\right)=5.3=15.1\)
Do đó x = 1 thì y = 15
x = 3 thì y =5
x = -15 thì y = -1
x = -3 thì y = -5
x = -5 thì y = -3
x = -1 thì y = -15
x = 5 thì y = 3
x = 15 thì y = 1
a) \(\frac{x}{7}+\frac{1}{14}=-\frac{1}{y}\)
\(\Rightarrow\frac{2x}{14}+\frac{1}{14}=\frac{-1}{y}\)
\(\Rightarrow\frac{2x+1}{14}=\frac{-1}{y}\)
\(\Rightarrow\left(2x+1\right).y=\left(-1\right).14=\left(-14\right)\)
Ta có bảng sau :
| 2x + 1 | 1 | -1 | 14 | -14 | 2 | -2 | 7 | -7 |
| 2x | 0 | -2 | 13 | -15 | 1 | -3 | 6 | -8 |
| x | 0 | -1 | \(\frac{13}{2}\) | \(\frac{-15}{2}\) | \(\frac{1}{2}\) | \(\frac{-3}{2}\) | 3 | -4 |
| y | -14 | 14 | -1 | 1 | -7 | 7 | -2 | 2 |
Vậy \(\left(x;y\right)\in\left\{\left(-1;14\right),\left(3;-2\right),\left(0;-14\right),\left(-4;2\right)\right\}\)
b) \(\frac{x}{9}+-\frac{1}{6}=-\frac{1}{y}\)
\(\Rightarrow\frac{2x}{18}+\frac{-3}{18}=\frac{-1}{y}\)
\(\Rightarrow\frac{2x-3}{18}=\frac{-1}{y}\)
\(\Rightarrow\left(2x-3\right).y=\left(-1\right).18=\left(-18\right)\)
Ta có bảng :
| 2x - 3 | 1 | -1 | 18 | -18 | 3 | -3 | 6 | -6 | 9 | -9 | -2 | 2 | ||||
| 2x | 4 | 2 | 21 | -15 | 6 | 0 | 9 | -3 | 12 | -6 | 1 | 5 | ||||
| x | 2 | 1 | \(\frac{21}{2}\) | \(\frac{-15}{2}\) | 3 | 0 | \(\frac{9}{2}\) | \(\frac{-3}{2}\) | 6 | -3 | \(\frac{1}{2}\) | \(\frac{5}{2}\) | ||||
| y | -18 | 18 | -1 | 1 | -6 | 6 | -3 | 3 | -2 | 2 | 9 | -9 |
Vậy \(\left(x;y\right)\in\left\{\left(2;-18\right),\left(1;18\right),\left(3;-6\right),\left(0;6\right),\left(6;-2\right),\left(-3,2\right)\right\}\)
a)\(\frac{x-1}{-3}=\frac{4}{7}\)
\(\Leftrightarrow7x-7=-12\)
\(\Leftrightarrow7x=-12+7\)
\(\Leftrightarrow7x=-5\)
\(\Leftrightarrow x=\frac{-5}{7}\)
vì \(x\in Z\Rightarrow x\in\left\{\varnothing\right\}\)
b) \(\frac{2}{3}=\frac{y+1}{-9}\)
\(\Leftrightarrow3y+3=-18\)
\(\Leftrightarrow3y=-18-3\)
\(\Leftrightarrow3y=-21\)
\(\Leftrightarrow y=-7\)
hok tốt!!
Sau đó ta có: x(y-1)+(y-1)=5
(x+1)(y-1)=5=1.5=5.1=-1.(-5)=-5.(-1)
Tự xét lập bảng...
Vậy x=0 thì y=6
x=4 thì y=2
x=-2 thì y=-4
x=-6 thì y=0
\(\frac{x}{5}-\frac{1}{y+2}=\frac{1}{10}\)
\(\frac{1}{y+2}=\frac{x}{5}-\frac{1}{10}=\frac{2x}{10}-\frac{1}{10}=\frac{2x-1}{10}\)
\(\Rightarrow\left(y+2\right).\left(2x-1\right)=1.10=10\)
\(\Rightarrow2x-1\inƯ\left(10\right)\)
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: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1982}+\frac{1}{1984}+\frac{1}{1986}\)