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a) x+2x+...+50x =2550
x. [ 1+2+3+....+50]=2550
ta co :
so so hang cua day 1;2;3;4;...;50:
[50-1]:1+1=50
tong cua day tren la :
[50+1].50:2=1275
=> x.1275=2550
x=2550:1275
vay x=2
Ta có : X = 3 + 32 + 33 + 34 +...+ 3100
=> 3X = 32 + 33 + 34 +...+ 3101
=> 3X - X = 3101 - 3
=> 2X = 3101 - 3
=> 2X + 3 = 3101
=> y = 101
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
34.x+4 = 81x+3 <=> 34.x+4 = 33.x+9 <=> 4.x+4 = 3.x+9 <=> 4.x - 3.x = 9-4 <=> x=5
Mk chỉ làm bài tính tổng thôi nhé!!!
A= 1+2+2^2+2^3+...+2^50
A.2= 2+2^2+2^3+...+2^50+2^51
A.2-A= (2+2^2+2^3+...+2^50+2^51)-(1+2+2^2+2^3+2^4+...+2^50)
A= 2^51-1
Vậy A= 2^51-1
B= 5+5^2+5^3+5^4+5^5+...+5^200
B.5= 5^2+5^3+5^4+...+5^200+5^201
B.5-B=5^201-5
B.4= 5^201-5
B= (5^201-5):4
Vậy B= (5^201-5):4
\(x+3x+3^2x+3^3x+...+3^{49}x=3^{100}-3^{50}\)
\(\Leftrightarrow x\left(1+3+3^2+3^3+...+3^{49}\right)=3^{100}-3^{50}\)
Đặt \(A=1+3+3^2+3^3+...+3^{49}\)
\(\Leftrightarrow3A=3+3^2+3^3+3^4+3^{50}\)
\(\Leftrightarrow3A-A=2A=3^{50}-1\)
\(\Leftrightarrow A=\frac{3^{50}-1}{2}\)
\(\Rightarrow\frac{x\left(3^{50}-1\right)}{2}=3^{100}-3^{50}\)
\(\Leftrightarrow x\left(3^{50}-1\right)=2.3^{100}-2.3^{50}\)
\(\Leftrightarrow x=\frac{2.3^{100}-2.3^{50}}{3^{50}-1}\)
\(x+3x+3^2x+3^3x+...+3^{49}x=3^{100}-3^{50}\)
\(\Leftrightarrow x\left(1+3+3^2+...+3^{49}\right)=3^{100}-3^{50}\)
\(\Leftrightarrow x\left(\frac{3^{50}-1}{2}\right)=3^{100}-3^{50}\)
\(\Leftrightarrow x\left(3^{50}-1\right)=3^{102}-3^{52}\)
\(\Rightarrow x=\frac{3^{102}-3^{52}}{3^{50}-1}=\frac{3^{52}\left(3^{50}-1\right)}{3^{50}-1}=3^{52}\)