Question

1 ) Tìm các số nguyên x ; y sao cho \(\left(x-1\right)\left(xy-5\right)=5\)

2 ) Cho ba sooss a , b , c khác 0 thỏa mãn điều kiện : \(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)

Tính giá trị biểu thức \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)

Answer 1

Bài 1:

\(\left(x-1\right).\left(xy-5\right)=5\)

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-1\in Z\\xy-5\in Z\end{matrix}\right.\)

\(\Rightarrow x-1\inƯC\left(5\right);xy-5\inƯC\left(5\right)\)

\(\Rightarrow x-1\in\left\{\pm1;\pm5\right\};xy-5\in\left\{\pm1;\pm5\right\}.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\\xy-5=5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=5\\xy-5=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-1\\xy-5=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-5\\xy-5=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\2y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\6y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\0y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\-4y=4\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=6\\y=1\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=-4\\y=-1\end{matrix}\right.\left(TM\right)\end{matrix}\right.\)

Vậy cặp số nguyên \(\left(x;y\right)\) thỏa mãn đề bài là: \(\left(2;5\right),\left(6;1\right),\left(0;0\right),\left(-4;-1\right).\)

Chúc bạn học tốt!

Answer 2

Bài 1:

Ta có bảng sau:

\(x-1\)	-5	-1	1	5
\(xy-5\)	-1	-5	5	1
x	-4	0	2	6
y	-1	mọi y∈Z	5	1

Answer 3

Bài 2:

Ta có:

\(\frac{3a+b+c}{a}=\frac{3b+a+c}{b}=\frac{3c+a+b}{c}\)

\(\Rightarrow\frac{3a}{a}+\frac{b+c}{a}=\frac{3b}{b}+\frac{a+c}{b}=\frac{3c}{c}+\frac{a+b}{c}\)

\(\Rightarrow3+\frac{b+c}{a}=3+\frac{a+c}{b}=3+\frac{a+b}{c}\)

\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{2\left(a+b+c\right)}{a+b+c}\)

Xét a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

Thay vào P, ta có:

\(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

Xét a+b+c\(\ne\)0

\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Thay vào P, ta có:

\(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=2+2+2=6\)

Vậy \(\left[{}\begin{matrix}P=-3\\P=6\end{matrix}\right.\)