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tìm chữ số tận cùng thì dc nhưng tính thì math error bạn ơi
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\(3^n+3^{n+1}=108\)
\(3^n+3^n.3=108\)
\(3^n.\left(1+3\right)=108\)
\(3^n.4=108\)
\(3^n=108:4\)
\(3^n=27\)
\(3^n=3^3\)
Vậy: \(n=3\)
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Ta có :
\(103-57:\left[-2x\left(2x-1\right)^2-\left(-9\right)^0\right]=-106\)
\(\Leftrightarrow57:\left[-2x\left(2x-1\right)^2-1\right]=103-\left(-106\right)\)
\(\Leftrightarrow57:\left[-2x\left(2x-1\right)^2-1\right]=209\)
\(\Leftrightarrow-2x\left(2x-1\right)^2-1=57:209\)
\(\Leftrightarrow-2x\left(2x-1\right)^2-1=\frac{3}{11}\)
\(\Leftrightarrow-2x\left(2x-1\right)^2=\frac{3}{11}+1\)
\(\Leftrightarrow-2x\left(2x-1\right)^2=\frac{14}{11}\)
\(\Leftrightarrow x\left(2x-1\right)^2=\frac{14}{11}:\left(-2\right)\)
\(\Leftrightarrow x\left(2x-1\right)^2=-\frac{7}{11}\)
K biết tớ giải sai hay đề cậu sai mà nó k ra kq cậu ạ :((
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\(C=1+5^2+5^4+5^6+...+5^{200}\)
\(25C=25+25^2+25^3+...+25^{201}\)
\(C=1.25+25^2+...+25^{200}\)
\(24C=25^{201}-1\)
\(C=\frac{25^{201}-1}{24}\)
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Ta thấy :
2a=a+a (1)
a^2=a.a (2)
Từ (1) và (2);ta có:
a.a > a+a
Nên a^2 > 2a
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d) 80 – [130 – (12 – 4)2] = 80 - (130 - 64) = 80 - 66 = 14.
tích mik na
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Bài làm
5.22x+1 + 22x+3 = 288
<=> 5.22x.2 + 22x.23 = 288
<=> 10.22x + 22x.8 = 288
<=> 22x( 10 + 8 ) = 288
<=> 22x.18 = 288
<=> 22x = 16
<=> 22x = 24
<=> 2x = 4
<=> x = 2
Vậy x = 2
(2x+1)4 = 625
(2x+1)4 = 54
suy ra 2x+1=5
2x=5+1
2x=6
x=6:2
x=3
625=5^4
Suy ra 2x+1=5
2x =4
x =2
Nhớ k mình đó