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Ta có " (x - 5)7 = (x - 5)4
=> (x - 5)7 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)3 - 1] = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^3-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
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\(2^x=2\Rightarrow x=1\)
\(2^{2x+2}=8^2\Rightarrow2^{2x+2}=2^6\Rightarrow2x+2=6\)\(2x=6-2=3\Rightarrow x=3:2=\frac{3}{2}\)
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\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
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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
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\(A=1+5+5^2+..+5^{49}+5^{50}\)
\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)
\(5A-A=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)
\(4A=\left(5-5\right)+\left(5^2-5^2\right)+...+\left(5^{50}+5^{50}\right)+5^{51}-1\)
\(4A=0+0+...+0+5^{51}-1\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
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1+3+5+...+x=1600
=(x+1).[(x-1):2+1] /2 =1600
=(x+1).(x+1) /2 =1600
=(x+1)^2:2=40^2
=(x+1):2=40
=x+1=80
=x=79
Giải:
\(\left(2x-5\right)^3=8\)
\(\Leftrightarrow\left(2x-5\right)^3=2^3\)
Vì \(3=3\)
Nên \(2x-5=2\)
\(\Leftrightarrow2x=2+5\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\dfrac{7}{2}=3,5\)
Vậy \(x=3,5\).
Chúc bạn học tốt!
Bn xem lại đề đi, mk biết làm nhưng làm ko ra dc