5x+3 \(\times\) 54x - 52x \(\times\)53x+2 =100
Tìm x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(5^{2x+1}\cdot5^{2x+2}-5^x\cdot5^{3x+2}=100\)
\(\Leftrightarrow5^{4x+3}-5^{4x+2}=100\)
\(\Leftrightarrow625^x\left(5^3-5^2\right)=100\)
\(\Leftrightarrow625^x=1\)
hay x=0
\(\Rightarrow5^{2x+1+2x+2}-5^{x+3x+2}=100\\ \Rightarrow5^{4x+3}-5^{4x+2}=100\\ \Rightarrow5^{4x+2}\left(5-1\right)=100\\ \Rightarrow5^{4x+2}=25=5^2\\ \Rightarrow4x+2=2\Rightarrow x=0\)
1.
a) \(5x.5x.5x=\left(5x\right)^3.\)
b) \(x^1.x^2.....x^{2006}=x^{\frac{\left(2006+1\right).2006}{2}=}x^{2013021}.\)
c) \(x^1.x^4.x^7.....x^{100}=x^{\frac{\left(100+1\right).\left(\frac{100-1}{3}+1\right)}{2}}=x^{1717}.\)
d) \(x^2.x^5.x^8.....x^{2003}=x^{\frac{\left(2003+2\right).\left(\frac{2003-2}{3}+1\right)}{2}}=x^{669670}.\)
2.
\(2^x+80=3^y\)
Với \(x>0\Rightarrow2^x\) chẵn
Và 80 chẵn
\(\Rightarrow2^x+80\) chẵn.
Mà \(3^y\) lẻ
\(\Rightarrow x< 0.\)
Mà \(x\in N\)
\(\Rightarrow x=0.\)
\(\Rightarrow2^0+80=3^y\)
\(\Rightarrow1+80=3^y\)
\(\Rightarrow3^y=81\)
\(\Rightarrow3^y=3^4\)
\(\Rightarrow y=4.\)
Vậy \(\left(x;y\right)=\left(0;4\right).\)
Chúc bạn học tốt!
Bài 1 :
a) 3. ( 5x - 7 )2 = 507
3 . 5x = 507 -72
3.5x = 435
x = 435 : 3.5
x = 29
d) 52x + 3 - 22 = 112
52x + 3 = 112 + 22
52x + 3 = 134
x = 134 - 52 + 3
x = 79
a)\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
b) k hiểu đề
a) A= (5x-2).(x+1)-(x-3).(5x+1)-17(x+3)
=> A= 5x2+5x-2x-2-5x2-x+15x+3-17x-51
=> A= -50
b) B= (6x-5) × ( x+8) - (3x-1) × (2x+3) - 9(4x-3)
=> B= 6x2+48x-5x-40-6x2-9x+2x+3-36x+27
=> B= -10
c) C = x(x3 + x2 - 3x -2 ) - ( x2 -2 ) × ( x2+x -1 )
=> C= x4+x3-3x2-2x-x4+x3+3x2-2x-2
=> C= 2x3-4x-2
\(x\times x^2\times x^3\times x^4\times...\times x^{100}=x^{1+2+3+4+...+100}=x^{101\times500}=x^{5050}\)
Áp dụng công thức: Nhân 2 lũy thừa cùng cơ số.
Ta có:
\(x\times x^2\times x^3\times...\times x^{100}\)
\(=x^{1+2+3+...+100}\)
\(x=5050\)
\(x.x^2.x^3...x^{100}=x^{1+2+3+...+100}\)
Đặt \(3^{1+2+3+...+100}=3^A\)
Ta có:
\(A=1+2+3+...+100\)
\(\Rightarrow A=100+99+98+...+1\)
\(\Rightarrow A=\left(1+100\right)+\left(2+99\right)+\left(3+98\right)+...+\left(100+1\right)\) ( 50 cặp số )
\(\Rightarrow A=101+101+101+...+101\) ( 50 số 101 )
\(\Rightarrow A=101.50\)
\(\Rightarrow A=5050\)
\(\Rightarrow3^A=3^{5050}\)
Vậy \(x.x^2.x^3...x^{100}=x^{5050}\)
Dấu chấm thay cho dấu nhân nhé!
...