Câu 1:
b) Ta có: \(2n-3⋮n+1\)
\(\Leftrightarrow2n+2-5⋮n+1\)
Vì \(2n+2=2\left(n+1\right)⋮n+1\)
nên \(-5⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(-5\right)\)
\(\Leftrightarrow n+1\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow n\in\left\{0;-2;4;-6\right\}\)(tm)
Vậy: \(n\in\left\{0;-2;4;-6\right\}\)
Câu 5:
\(2x+\frac{1}{7}=\frac{1}{y}\)
\(\Rightarrow\frac{2x}{1}+\frac{1}{7}=\frac{1}{y}\)
\(\Rightarrow\frac{2x.7}{1.7}+\frac{1}{7}=\frac{1}{y}\)
\(\Rightarrow\frac{14x}{7}+\frac{1}{7}=\frac{1}{y}\)
\(\Rightarrow\frac{14x+1}{7}=\frac{1}{y}.\)
\(\Rightarrow\left(14x+1\right).y=1.7\)
\(\Rightarrow\left(14x+1\right).y=7\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}14x+1\in Z\\y\in Z\end{matrix}\right.\)
\(\Rightarrow14x+1\inƯC\left(7\right);y\inƯC\left(7\right)\)
\(\Rightarrow14x+1\in\left\{1;-1;7;-7\right\};y\in\left\{1;-1;7;-7\right\}.\)
Đến đoạn này thì bạn tự lập bảng ra nhé, cái nào là số nguyên thì lấy.
Chúc bạn học tốt!
Câu 2:
a) Ta có: \(3x-\left|2x+1\right|=2\)
\(\Leftrightarrow\left|2x+1\right|=3x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-2\\2x+1=2-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+1-3x+2=0\\2x+1-2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3-x=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{3;\frac{1}{5}\right\}\)
Câu 3:
Gọi ba phân số cần tìm là \(\frac{a}{x}\); \(\frac{b}{y}\); \(\frac{c}{z}\)(x,y,z≠0)
Theo đề bài, ta có:
\(a:b:c=3:4:5\)
⇔a=3m; b=4m; c=5m
\(x:y:z=5:1:2\)
⇔x=5n; y=n; z=2n
Và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{213}{70}\)
⇒\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{3m}{5n}+\frac{4m}{n}+\frac{5m}{2n}=\frac{213}{70}\)
⇔\(\frac{3}{5}\cdot\frac{m}{n}+4\cdot\frac{m}{n}+\frac{5}{2}\cdot\frac{m}{n}=\frac{213}{70}\)
⇔\(\frac{m}{n}\cdot\left(\frac{3}{5}+4+\frac{5}{2}\right)=\frac{213}{70}\)
⇔\(\frac{m}{n}=\frac{213}{70}:\frac{71}{10}=\frac{213}{70}\cdot\frac{10}{71}=\frac{3}{7}\)
Ta có: \(\frac{a}{x}=\frac{m}{n}\cdot\frac{3}{5}\)
hay \(\frac{a}{x}=\frac{3}{7}\cdot\frac{3}{5}=\frac{9}{35}\)
Ta có: \(\frac{b}{y}=4\cdot\frac{m}{n}\)
hay \(\frac{b}{y}=4\cdot\frac{3}{7}=\frac{12}{7}\)
Ta có: \(\frac{c}{z}=\frac{5}{2}\cdot\frac{m}{n}\)
hay \(\frac{c}{z}=\frac{5}{2}\cdot\frac{3}{7}=\frac{15}{14}\)
Vậy: Ba phân số cần tìm lần lượt là: \(\frac{9}{35}\); \(\frac{12}{7}\); \(\frac{15}{14}\)
Bài 2:
b)
Ta có: \(3\left(x-1\right)=2\left(y-2\right)\)
⇔\(\frac{x-1}{2}=\frac{y-2}{3}\)(1)
Ta có: \(4\left(y-2\right)=3\left(z-3\right)\)
⇔\(\frac{y-2}{3}=\frac{z-3}{4}\)(2)
Từ (1) và (2) suy ra
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}\)
Ta có: \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\) và 2x+3y-z=50
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)
Do đó:
\(\left\{{}\begin{matrix}\frac{2x-2}{4}=5\\\frac{3y-6}{9}=5\\\frac{z-3}{4}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2=20\\3y-6=45\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=22\\3y=51\\z=23\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Vậy: x=11; y=17 và z=23
Bài 1:
b) https://olm.vn/hoi-dap/detail/33227335613.html
