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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)
25x + 3 . (4-6x) =50
=> 25x +12-18x=50
=> 25x-18x=50-12
=> 7x=38
=> x=38/7
Vay x=38/7
25x .75 x = 0
( 25 x 75 ) . ( x .x ) = 0
1875 . x2 = 0
=> x = 0
a/ 25x = x <=> 25x - x = 0 => 24x = 0 => x = 0
b/ (x - 2016) . (x - 2018) = 0
=> x - 2016 = 0 hoặc x - 2018 = 0
=> x = 2016 hoặc x = 2018
\(b,\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{1}{7}\)
\(\dfrac{1}{5}:x=\dfrac{7}{35}-\dfrac{5}{35}\)
\(\dfrac{1}{5}:x=\dfrac{2}{35}\)
\(x=\dfrac{1}{5}:\dfrac{2}{35}\)
\(x=\dfrac{1}{5}.\dfrac{35}{2}\)
\(x=\dfrac{7}{2}\)
\(\text{Cx khá đơn giản nhưng a giải chi tiết luôn nha}\)
\(1.\)
\(5\left(x+4\right)=25x\)
\(\Leftrightarrow5x+20=25x\)
\(\Leftrightarrow5x-25x=-20\)\(\text{(chuyển vế)}\)
\(\Leftrightarrow-20x=-20\)
\(\Leftrightarrow x=1\)
\(2.\)
\(-\frac{3}{6}=\frac{25}{2x-1}\)
\(\Leftrightarrow-\frac{1}{2}=25.\frac{1}{2x-1}\)
\(\Leftrightarrow\frac{1}{2x-1}=-\frac{1}{2}:25=-\frac{1}{50}\)
\(3.\)\(\text{(tương tự)}\)
\(\Leftrightarrow-50=2x-1\)\(\text{(nhân chéo)}\)
\(\Leftrightarrow x=-50+1=-49\)
\(4.\)
\(5\left(x-4\right)=3x+8\)
\(\Leftrightarrow5x-20=3x+8\)
\(\Leftrightarrow5x-3x=20+8\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=28:2=14\)
1) 5.(x+4)=25x
5x+20=25x
20=25x-5x
20=20x
x=1
2) \(\frac{-3}{6}=\frac{25}{2x-1}\)
-3.(2x-1)=6.25
-6x+3=150
-6x=147
x=-24,5
câu 3,4 lm tương tự nha Nhớ đấy!!!!
\(x^3-\frac{4}{25}x=0\)
\(\Rightarrow x\left(x^2-\frac{4}{25}\right)=0\)
\(\Rightarrow x=0\)hoặc \(x^2-\frac{4}{25}=0\)
Nếu \(x^2-\frac{4}{25}=0\)
\(\Rightarrow x^2=0+\frac{4}{25}\)
\(\Rightarrow x^2=\frac{4}{25}\)
=>\(x\in\left\{\frac{-2}{45};\frac{2}{45}\right\}\)
Vậy \(x\in\left\{\frac{-2}{45};\frac{2}{45};0\right\}\)