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Để A= \(\frac{5n+1}{n+1}\)
thì \(5n+1\)chia hết cho n +1 nên n+1 thuộc U(5)=1, 5.-1,-5
Ta có
Nếu n+1 =1 thì suy ra n =0
....n+1 = -1 thì suy ra n= -2
... n+1=5 thì suy ra n =4
....n+1= -5 thì suy ra n = -6
vây n thuộc 0, -2, 4, -6
Để \(A=\frac{5n+1}{n+1}\in Z\) \(\Leftrightarrow5n+1⋮n+1\)
\(\Leftrightarrow\) \(5n+1-5\left(n+1\right)⋮n+1\) (Vì 5(n+1)⋮n+1)
\(\Leftrightarrow5n+1-5n-5⋮n+1\)
\(\Leftrightarrow-4⋮n+1\)
\(\Rightarrow n+1\in\) Ư\(\left(-4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow n\in\left\{0;1;3;-2;-3;-5\right\}\)
Mà \(n\in N\) nên \(n\in\left\{0;1;3\right\}\)
Vậy để \(A\) nguyên thì \(n\in\left\{0;1;3\right\}\) (\(n\in N\))
2: \(A=9^n\cdot81-9^n+3^n\cdot9+3^n\)
\(=9^n\cdot80+3^n\cdot10\)
\(=10\left(9^n\cdot8+3^n\right)⋮10\)
a) (12)m=132(12)m=132
\(\Rightarrow\left(\dfrac{1}{2}\right)^m=\left(\dfrac{1}{2}\right)^5\Rightarrow m=5\)
b)
343125=(75)n
\(\Rightarrow\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\Rightarrow n=3\)
b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Đề thiếu điều kiện \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) nữa đấy
Ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\)
\(=\dfrac{a+b+c}{a+b+c}\)
\(=1\)
Với \(\dfrac{a+b-c}{c}=1\)
\(\Rightarrow a+b-c=c\)
\(\Rightarrow a+b=2c\left(1\right)\)
Với \(\dfrac{b+c-a}{a}=1\)
\(\Rightarrow b+c-a=a\)
\(\Rightarrow b+c=2a\left(2\right)\)
Với \(\dfrac{c+a-b}{b}=1\)
\(\Rightarrow c+a-b=b\)
\(\Rightarrow c+a=2b\left(3\right)\)
Ta lại có:
\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{b}{b}+\dfrac{a}{b}\right)\left(\dfrac{c}{c}+\dfrac{b}{c}\right)\left(\dfrac{a}{a}+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)
Thay (1) , (2) và (3) vào ta được
\(=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\)
\(=\dfrac{8abc}{abc}\)
\(=8\)
1: \(=\left|\dfrac{-21+5}{35}\right|+\dfrac{5}{7}\cdot\dfrac{-2}{5}\)
\(=\dfrac{16}{35}+\dfrac{-2}{7}=\dfrac{16}{35}-\dfrac{10}{35}=\dfrac{6}{35}\)
2: =>4^x+1=16
=>x+1=2
=>x=1
4/ \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{24}\end{matrix}\right.\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\) (đặt k)
Suy ra \(x=15k;y=20k;z=24k\)
Thay vào,ta có:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
$A=\frac{5n+1}{n+1}=\frac{5(n+1)-4}{n+1}=5-\frac{4}{n+1}\in \mathbb{Z}$
$\Leftrightarrow n+1\in Ư(4)=\left\{-4;-2;-1;1;2;4\right\}$
Mà $n\in\mathbb{N}$
$\Rightarrow n\in\left\{0;1;3\right\}$
\(A=\dfrac{5n+1}{n+1}=\dfrac{5\left(n+1\right)-4}{n+1}=\dfrac{5\left(n+1\right)}{n+1}-\dfrac{4}{n+1}=5-\dfrac{4}{n+1}\).ĐK:n≠-1
để \(Anguy\text{ê}n.th\text{ì}4⋮(n+1)\\ \Rightarrow n+1\in\text{Ư}\left(4\right)=\left\{1;2;4\right\}\)
ta có bảng sau :
vậy....