Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 71.2-6(2x+5)=105:103
<=> 71.2-6(2x+5)=102
142-6(2x+5)=100
6(2x+5)=142-100
6(2x+5)=42
2x+5=7
2x=7-5
2x=2
x=2:2
x=1
a,71.2-6.(2x+5)=10^5:10^3
142-6.(2x+5)=10^2
142-6.(2x+5)=100
6.(2x+5)=142-100
6.(2x+5)=42
2x+5=42:6
2x+5=7
2x=7-5
2x=2
x=1
Vậy x=1
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
\(\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+....+\left(2^2+2\right)\)
\(=2^9.\left(2+1\right)+2^7.\left(2+1\right)+...+2.\left(2+1\right)\)
\(=2^9.3+2^7.3+...+2.3\)
\(=3.\left(2^9+2^7+...+2\right)⋮3\)
P/S: mấy bài khác tương tự
\(a,2^{10}+2^9+2^8+...+2\)
\(=\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+...+\left(2^2+2\right)\)
\(=2^9\left(2+1\right)+2^7\left(2+1\right)+...+2\left(2+1\right)\)
\(=2^9.3+2^7.3+...+2.3\)
\(=3\left(2^9+2^7+...+2\right)⋮3\left(đpcm\right)\)
\(b,1+3+3^2+3^3+...+3^{99}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)
\(=4+3^2\left(1+3\right)+...+3^{98}\left(1+3\right)\)
\(=4+3^2.4+...+3^{98}.4\)
\(=4\left(1+3^2+...+3^{98}\right)⋮4\left(đpcm\right)\)
\(c,1+5+5^2+5^3+...+5^{1975}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{1974}+5^{1975}\right)\)
\(=6+5^2\left(1+5\right)+...+5^{1974}\left(1+5\right)\)
\(=6+5^2.6+...+5^{1974}.6\)
\(=6\left(1+5^2+...+5^{1974}\right)⋮6\left(đpcm\right)\)
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 \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a, Ta có: \(142-\left(12x+30\right)=10^{5-3}\)
\(=>142-12x-30=10^2\)
\(112-12x=100\)
\(=>12x=112-100=12=>x=1\)
Vậy số cần tìm là 1;
b, Ta có: \(=>\left(5x+3^4\right)=6^9:6^8.3^4\)
\(=>5x+3^4=6.3^4=>5x=6.3^4-3^4\)
\(=>5x=5.3^4=>x=3^4=81\)
Vậy x=81;
CHÚC BẠN HỌC TỐT.......
Ta có: \(71.2-6(2x+5)=10^5:10^3\)
\(\Rightarrow142-12x-30=10^2\)
\(\Rightarrow142-30-10^2=12x\)
\(\Rightarrow142-30-100=12x\)
\(\Rightarrow12=12x\)
\(\Rightarrow x=\dfrac{12}{12}\)
\(\Rightarrow x=1\)
Vậy \(x=1\)