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Bài 1:
a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)
\(xy\) = 12
12 = 22.3; Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6;12}
Lập bảng ta có:
| \(x\) | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| y | -1 | -2 | -3 | -4 | -6 | -12 | 12 | 6 | 4 | 3 | 2 | 1 |
Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)
b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)
\(x\) = \(\dfrac{2}{7}\).y
\(x\) \(\in\)z ⇔ y ⋮ 7
y = 7k;
\(x\) = 2k
Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\)
\(\frac{-4}{8}=\frac{x}{-10}=\frac{-7}{y}=\frac{z}{-24}\)
\(\Rightarrow\frac{-4}{8}=\frac{x}{-10}\Leftrightarrow x=\frac{-10.\left(-4\right)}{8}=5\)
\(\Rightarrow\frac{-4}{8}=\frac{-7}{y}\Leftrightarrow y=\frac{-7.8}{-4}=14\)
\(\Rightarrow\frac{-4}{8}=\frac{z}{-24}\Leftrightarrow z=\frac{-24.\left(-4\right)}{8}=12\)
Vậy
\(\frac{x}{2}=\frac{8}{x}\)
\(\Rightarrow x.x=2.8\)
\(x^2=16\)
\(x^2=\left(\pm4\right)^2\)
\(\Rightarrow x=\pm4\)
học tốt
c)\(-\frac{4}{8}=\frac{x}{-10}=-\frac{7}{y}=\frac{z}{-24}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{4}{8}=\frac{x}{-10}\\-\frac{4}{8}=-\frac{7}{y}\\-\frac{4}{8}=\frac{z}{-24}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\left(-4\right).\left(-10\right):8=5\\y=8.\left(-7\right):\left(-4\right)=14\\z=-4.\left(-24\right):8=12\end{cases}}}\)
vậy x=5;y=14;z=12
d) \(\frac{x}{2}=\frac{8}{x}\)
\(\Leftrightarrow x^2=2.8\)
\(\Leftrightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
\(\frac{-4}{8}=\frac{x}{-10}=\frac{-7}{y}=\frac{z}{-24}\Rightarrow\frac{-1}{2}=\frac{5}{-10}=-\frac{7}{14}=\frac{12}{-24}\Rightarrow x=5;y=14;z=12\)
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
\(\frac{x}{-2}=-\frac{8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
\(-\frac{2}{x}=\frac{y}{3}\)
\(\Leftrightarrow xy=-6\)
\(\Leftrightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Xét bảng
| x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -1 |
| y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 6 |
Vậy.................
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Leftrightarrow2x-9=\frac{240.39}{80}\)
\(\Leftrightarrow2x-9=117\)
\(\Leftrightarrow2x=126\)
\(\Leftrightarrow x=63\)
a, \(\frac{17}{y}=\frac{-7}{11}\)
\(\Rightarrow17\cdot11=-7\cdot y\)
\(\Rightarrow187=-7\cdot y\)
\(\Rightarrow\frac{187}{-7}=y\)
b, \(\frac{-8}{3x-1}=\frac{4}{7}\)
\(\Rightarrow\frac{-8}{3x-1}=\frac{-8}{-14}\)
\(\Rightarrow3x-1=-14\)
\(\Rightarrow3x=-14+1\)
\(\Rightarrow3x=-13\)
\(\Rightarrow x=\frac{-13}{3}\)
c, \(\frac{x}{-3}=\frac{-3}{x}\)
\(\Rightarrow x\cdot x=-3\cdot\left(-3\right)\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x^2=\left(\pm3\right)^2\)
\(\Rightarrow x=\pm3\)
d, \(\frac{-4}{y}=\frac{x}{2}\)
\(\Rightarrow-4\cdot2=x\cdot y\)
\(\Rightarrow-8=x\cdot y\)
\(\Rightarrow x;y\inƯ\left(-8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
ta có bảng :
| x | -1 | -8 | -2 | -4 |
| y | 8 | 1 | 4 | 2 |
a)\(\frac{14}{y}\)\(=\) \(\frac{-7}{11}\)
\(\Rightarrow\)\(14\cdot11=y\cdot\left(-7\right)\)
\(y=\)\(\frac{14\cdot11}{-7}\)
\(y=22\)
c) \(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(\Rightarrow\) \(x\cdot x=\left(-3\right)\cdot\left(-3\right)\)
\(\Rightarrow\)\(x^2=9\)
\(\Rightarrow\)\(x^2=9\)hoặc \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)
a./ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}=\frac{x-y+z}{5-7+4}=\frac{-10}{2}=-5\)
\(\Rightarrow x=-25;y=-35;z=-20\)
b./ \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}=\frac{x+y-z}{5-4-\left(-7\right)}=\frac{-40}{6}=-5\)
\(\Rightarrow x=-25;y=20;z=35\)
đề yêu cầu gì vậy