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\(\frac{x+1+3}{x+1}=\frac{x+1}{x+1}+\frac{3}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Ta có :
\(x+1+3⋮x+1\)
=> \(x+1+3-\left(x+1\right)⋮x+1\)
=> \(3⋮x+1\)
=> \(x+1\inƯ_3\)
=> \(x+1\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{0;2;-2;-4\right\}\)
Giải:
Ta có:
\(x+8⋮x+1\)
\(\Rightarrow\left(x+1\right)+7⋮x+1\)
\(\Rightarrow7⋮x+1\)
\(\Rightarrow x+1\in\left\{1;-1;7;-7\right\}\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
+) \(x+1=7\Rightarrow x=6\)
+) \(x+1=-7\Rightarrow x=-8\)
Vậy \(x\in\left\{0;-2;6;-8\right\}\)
\(\frac{x+8}{x+1}=\frac{x+7+1}{x+1}=\frac{x+1}{x+1}+\frac{7}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(7\right)=\left\{1;7;-1;-7\right\}\)
\(\Rightarrow x\in\left\{0;6;-2;-8\right\}\)
\(\frac{x+4}{x}=\frac{x}{x}+\frac{4}{x}\)
\(\Rightarrow x\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
Ta có :
\(x+4⋮x\)
=> \(x+4-x⋮x\)
=> \(4⋮x\)
=> \(x\inƯ_4\)
=> x\(\in\left\{1;2;4;-1;-2;-4\right\}\)
Ta có :
\(9⋮2x+1\)
=> \(2x+1\inƯ_9\)
=> \(2x+1\in\left\{1;3;9\right\}\)
=> \(2x\in\left\{0;2;8\right\}\)
=> \(x\in\left\{0;1;4\right\}\)
Vậy \(x\in\left\{0;1;4\right\}\)
3x+2=369
=>x+2=69
x=69-2
x=67
2x-5=810
2x-5=230
=>x-5=30
x=30+5
x=35
3x+2+3x=810
3x.32+3x=810
3x.(32+1)=810
3x.10=810
3x=810:10
3x=81
3x=34
=>x=4
5x+1-5x=500
5x.5-5x=500
5x.(5-1)=500
5x.4=500
5x=500:4
5x=125
5x=53
=>x=3
a) 3x+2 = 369
x + 2 = 69
x = 69 - 2
x = 67
b) 2x-5 = 810
2x-5 = 230
x - 5 = 30
x = 30 + 5
x = 35
c) 3x+2 + 3x = 810
3x . 9 + 3x . 1 = 810
3x . ( 9 + 1 ) = 810
3x . 10 = 810
3x = 810 : 10
3x = 81
3x = 34
=> x = 4
d) 5x+1 - 5x = 500
5x . 5 - 5x . 1 = 500
5x . ( 5 - 1 ) = 500
5x . 4 = 500
5x = 500 : 4
5x = 125
5x = 53
=> x = 3
a) \(3^{x+1}=729\)\(\Leftrightarrow3.3^x=729\Rightarrow3^x=243\Rightarrow3^x=3^5\Rightarrow x=5\)
b) \(2^x+2^{x+1}+2^{x+3}=704\Rightarrow2^x+2^x.2+2^x.2^3=704\Rightarrow2^x\left(1+2+8\right)=704\)
\(\Rightarrow2^x.11=704\Rightarrow2^x=64\Rightarrow2^x=2^5\Rightarrow x=5\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
\(\frac{x+5}{x+1}=\frac{x+4+1}{x+1}=\frac{x+1}{x+1}+\frac{4}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{0;1;3;-2;-3;-5\right\}\)
x+5\(⋮\)x+1
x+1+4\(⋮\)x+1
Vì x+1\(⋮\)x+1
Buộc 4\(⋮\)x+1=>x+1ϵƯ(4)={1;2;4}
Với x+1=1=>x=0
x+1=2=>x=1
x+1=4=>x=3
Vậy xϵ{0;1;3}