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Ta có:
\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot5^n\)
\(2^n\left(\frac{1}{2}+4\right)=9\cdot5^n\)
\(\frac{9}{2}\cdot2^n=9\cdot5^n\)
Tức: \(9\cdot\frac{1}{2}\cdot2^n=9\cdot5^n\)
Suy ra: \(2^{n-1}=5^n\)
Nhận thấy: \(n-1< n\)
Hơn nữa \(2< 5\)
Do đó: \(2^{n-1}< 5^n\)
Vậy không có n thỏa mãn
1: \(\Leftrightarrow\left(\dfrac{2}{5}x+2\right)=-\dfrac{3}{2}\cdot\left(-4\right)=6\)
=>2/5x=4
hay x=10
2: \(x\left(x+1\right)=0\)
=>x=0 hoặc x+1=0
=>x=0 hoặc x=-1
3: \(2x-16=x+40\)
=>2x-x=40+16
=>x=56
4: \(\dfrac{x}{4}=\dfrac{9}{x}\)
nên \(x^2=36\)
=>x=6 hoặc x=-6
\(\frac{2}{n}+\frac{2}{n+1}=\frac{2\left(n+1\right)}{n\left(n+1\right)}+\frac{2n}{n\left(n+1\right)}\)\(=\frac{2\left(n+1\right)+2n}{n\left(n+1\right)}=\frac{2n+2+2n}{n\left(n+1\right)}=\frac{4n+2}{n\left(n+1\right)}\)
\(\frac{1}{n\left(n+1\right)}+\frac{-2}{n+1}=\frac{1}{n\left(n+1\right)}+\frac{-2n}{n\left(n+1\right)}\)\(=\frac{1+\left(-2n\right)}{n\left(n+1\right)}=\frac{1-2n}{n\left(n+1\right)}\)
\(7-5\left(x-2\right)=3+2\left(4-x\right)\)
\(\Leftrightarrow7-5x+10=3+8-2x\)
\(\Leftrightarrow17+5x=11-2x\)
\(\Leftrightarrow17-11=-2x-5x\)
\(\Leftrightarrow6=-7x\)
\(\Leftrightarrow x=\frac{-7}{6}\)
7 - 5( x - 2) = 3 + 2 (4-x )
=> 7 - 5x + 10 = 3 + 8 - 2x
=> 17 + 5x = 11 - 2x
=> 17 - 11 = -2x - 5x
=>6 = -7x
=> x = -7/6
Vay x = -7/6
Chuc ban hc tot
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
a. Vì A thuộc Z
\(\Rightarrow x-2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-3;1;3;7\right\}\)( tm x thuộc Z )
b. Ta có : \(B=\frac{x+2}{x-3}=\frac{x-3+5}{x-3}=1+\frac{5}{x-3}\)
Vì B thuộc Z nên 5 / x - 3 thuộc Z
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-2;2;4;8\right\}\)( tm x thuộc Z )
c. Ta có : \(C=\frac{x^2-x}{x+1}=\frac{x^2+x-2x+2-2}{x+1}=\frac{x\left(x+1\right)-2x+2-2}{x+1}\)
\(=x-2-\frac{2}{x+1}\)
Vi C thuộc Z nên 2 / x + 1 thuộc Z
\(\Rightarrow x+1\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x\in\left\{-3;-2;0;1\right\}\) ( tm x thuộc Z )
3B=3^1+3^2+3^3+.....+3^119+3^120
3B-B=(3^1+3^2+3^3+.....+3^119+3^120)-(1+3^1+3^2+3^3+.....+3^119)
2B=3^120-1
B=3^120-1/2
\(B=1+3^1+3^2+...+3^{118}+3^{119}\)
\(3B=3+3^2+3^3+..+3^{120}\)
\(3B-B=\left(3+3^2+...+3^{120}\right)-\left(1+3+3^2+...+3^{119}\right)\)
\(2B=1+3^{120}\)
Bài làm:
Ta có:
a) \(x^{n+1}\div5=5^n\)
\(\Leftrightarrow x^{n+1}=5^n.5\)
\(\Leftrightarrow x^{n+1}=5^{n+1}\)
\(\Rightarrow x=5\)
b) \(x^n.9=3^{n+2}\)
\(\Leftrightarrow x^n.3^2=3^{n+2}\)
\(\Leftrightarrow x^n=3^{n+2}\div3^2\)
\(\Leftrightarrow x^n=3^n\)
\(\Rightarrow x=3\)