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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 \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
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 :
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 \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)