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1)\(25x+3\left(4-6x\right)=50\)
\(25x+12-18x=50\)
\(7x+12=50\)
\(7x=38\)
\(x=\frac{38}{7}\)
2)\(4\left(2x+3\right)+2\left(3x+1\right)=120\)
\(8x+12+6x+2=120\)
\(14x+14=120\)
\(14x=106\)
\(x=\frac{53}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
a) A = 2^0 + 2^1 + 2^2 +...+ 2^50
2A=2^1+2^2+2^3+...+2^51
2A-A=(2^1+2^2+2^3+...+2^51)-(2^0 + 2^1 + 2^2 +...+ 2^50)
A=(2^1-2^1)+(2^2-2^2)+...+(2^50-2^50)+(2^51-2^1)
A=0+0+...+0+(2^51-2^1)
A=2^51-2^1
b)B = 5 + 5^2 + 5^3 +...+ 5^99 + 5^100
5B=5^2+5^3+5^4+...+5^100+5^101
5B-B=(5^2+5^3+5^4+...+5^100+5^101)-( 5 + 5^2 + 5^3 +...+ 5^99 + 5^100)
4B=(5^2-5^2)+(5^3-5^3)+...+(5^100-5^100)+(5^101-5)
4B=0+0+...+0+(5^101-5)
4B=5^101-5
B=(5^101-5)/4
c)C = 3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010
3C=3^2-3^3+3^4-3^5+...+3^2010-3^2011
3C-C=(3^2-3^3+3^4-3^5+...+3^2010-3^2011)-(3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010)
...............................................!!!!!!!!!!!!!!!!!!!!!!!!
Bài 2
8(mình k0 chắc)
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Câu 2
\(20-\left(7\left(x-3\right)+4\right)=-\left(7x-37\right)\)
\(-\left(7x-37\right)=2\)
\(37-7x=2\)
\(-7x=-35\)
\(\Rightarrow x=5\)
Câu 1:
\(x\in\left(\infty,-\infty\right)\)
\(1.3x.32=96x\)
\(96x=3^5\)
\(96x=243\)
\(32x=81\)
\(\Rightarrow x=2\frac{17}{32}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 219-7x-7=100
212-7x=100
7x=112
x=16
các câu còn lại cứ thế mà làm
25x + 3 . (4-6x) =50
=> 25x +12-18x=50
=> 25x-18x=50-12
=> 7x=38
=> x=38/7
Vay x=38/7