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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/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
2x - 138 = 23 . 32
=> 2x - 138 = 72
=> 2x = 72 + 138
=> 2x = 210
=> x = 105
10 + 2x = 43 . 42
10 + 2x =1024
2x = 1024 - 10
2x = 1014
x = 507
(3x - 24) . 73 = 2.74
3x - 24 = 2 . 74 : 73
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 10
x4 = 16
=> x4 = 24
=> x = 2
a, \(x^5-x^2=0\)
\(\Rightarrow x^2\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^3=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
b, \(x^{2020}-x^{2019}=0\)
\(\Rightarrow x^{2019}\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{2019}=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
c, \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^4-1\right]\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^4-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=6\end{cases}}}\)
a) \(x^5-x^2=0\)
\(\Rightarrow x^2\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^3-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^3=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x\in\left\{0;1\right\}\)
b) \(x^{2020}-x^{2019}=0\)
\(\Rightarrow x^{2019}\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{2019}=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x\in\left\{0;1\right\}\)
Câu c tương tự nhé em!
Chúc em học tốt nhé!
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
5A=5+52+53+...+5201
5A-A=(5+52+53+...+5201)-(1+5+52+53+...+5200)=5201-1
=> A=\(\frac{5^{201}-1}{4}\)
mk ko bt bn lm sai hay đúng nhưng kết quả đúng là 201 cơ, nhưng dù sao cx pải k cho bn, thank đã giúp mk nha ( mặc dù mk lm xong lâu rùi )
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62