Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. 2n-3 ⋮ n+1
⇒2n+2-5 ⋮ n+1
⇒2(n+1)-5 ⋮ n+1
Do n∈Z
⇒n+1 ∈ Ư(-5)={-1,1,-5,5}
⇒\(\left[{}\begin{matrix}n-1=-1\\n-1=1\\n-1=-5\\n-1=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=2\\n=-4\\n=6\end{matrix}\right.\)
Vậy x∈{0,2,-4,6}
2. Ta có:
x-y-z=0 ⇒\(\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\)
Thay vào biểu thức ta được:
\(B=\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)
⇒\(B=\frac{x-x+y}{x}.\frac{y-y-z}{y}.\frac{z+x-z}{z}\)
⇒\(B=\frac{y.\left(-z\right).x}{x.y.z}=\frac{\left(-1\right)xyz}{xyz}=-1\)
Vậy biểu thức B có giá trị là -1
1.
\(10x=|x+\dfrac{1}{10}|+|x+\dfrac{2}{10}|+...+|x+\dfrac{9}{10}| \ge 0\)
\(\Rightarrow x\ge0\)
\(pt\Leftrightarrow x+\frac{1}{10}+x+\frac{2}{10}+...+x+\frac{9}{10}=10x\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{2}{10}+...+\frac{9}{10}=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}\)
4.
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{b+3c}=\frac{b}{c+3a}=\frac{c}{a+3b}=\frac{a+b+c}{4\left(a+b+c\right)}=\frac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}4a=b+3c\left(1\right)\\4b=c+3a\left(2\right)\\4c=a+3b\left(3\right)\end{matrix}\right.\)
Từ \(\left(1\right);\left(2\right)\Rightarrow4a=b+3\left(4b-3a\right)\)
\(\Rightarrow12a=12b\Rightarrow a=b\left(4\right)\)
Từ \(\left(1\right);\left(3\right)\Rightarrow4c=a+3\left(4a-3c\right)\)
\(\Rightarrow12a=12c\Rightarrow a=c\left(5\right)\)
Từ \(\left(4\right);\left(5\right)\Rightarrow a=b=c\left(đpcm\right)\)
1) a. Ta có:\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
\(\Rightarrow\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)
\(\Rightarrow\frac{x+4+2008}{2008}+\frac{x+3+2009}{2009}=\frac{x+2+2010}{2010}+\frac{x+1+2011}{2011}\)
\(\Rightarrow\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)
\(\Rightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)=\left(x+2012\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)\)
\(\Rightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)-\left(x+2012\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)=0\)
\(\Rightarrow\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
\(\Rightarrow x+2012=0\)
\(\Rightarrow x=-2012\)
Bài 2:
a.Ta có: \(\frac{x+2y}{18}=\frac{1+4y}{24}\)
\(\Rightarrow24x+48y=18+72y\)
\(\Rightarrow24x+48y-72y=18\)
\(\Rightarrow24x-24y=18\)
\(\Rightarrow24\left(x-y\right)=18\)
\(\Rightarrow x-y=\frac{3}{4}\)
\(\Rightarrow y=x-\frac{3}{4}\)
thay \(y=x-\frac{3}{4}\)vào \(\frac{1+4y}{24}=\frac{1+x+6y}{6x}\)ta được \(\frac{1+4\times\left(x-\frac{3}{4}\right)}{24}=\frac{1+x+6\times\left(x-\frac{3}{4}\right)}{6x}\)
giải ra ta được x=7
\(\Rightarrow y=7-\frac{3}{4}=\frac{25}{4}\)
b. Đẻ A mang giá trị nuyên
\(\Leftrightarrow9+3n⋮n-4\)
\(\Leftrightarrow3n-12+21⋮n-4\)
\(\Leftrightarrow3\left(n-4\right)+21⋮n-4\)
\(\Leftrightarrow21⋮n-4\)
\(\Leftrightarrow n-4\inƯ_{\left(21\right)}=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Ta có bảng sau:
| n-4 | 1 | -1 | 3 | -3 | 7 | -7 | 21 |
-21 |
| n | 5 | 4 | 7 | 1 | 11 | -3 | 25 | -17 |
Vậy \(n\in\left\{5;4;7;1;11;-3;25;-17\right\}\)thì A là số nguyên.
Thay n vào A và tính giá trị