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\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
Bài 3
\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)
\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)
\(=\dfrac{33}{23}\le x\le3\)
Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)
Có 1 giá trị thỏa mãn
Chọn A
Bài 4
\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)
Chọn D
Bài 5
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{100}{100}-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
CHọn C
a, \(x-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x-1 | 1 | -1 | 3 | -3 |
| x | 2 | 0 | 4 | -2 |
b, \(2x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| 2x-1 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 1 | 0 | loại | loại | loại | loại |
c, \(\dfrac{3\left(x-1\right)+10}{x-1}=3+\dfrac{10}{x-1}\Rightarrow x-1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
| x-1 | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
| x | 2 | 0 | 3 | -1 | 6 | -4 | 11 | -9 |
d, \(\dfrac{4\left(x-3\right)+3}{-\left(x-3\right)}=-4-\dfrac{3}{x+3}\Rightarrow x+3\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x+3 | 1 | -1 | 3 | -3 |
| x | -2 | -4 | 0 | -6 |
