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\(A,\left(a^6\right)^4.a^{12}=a^{24}.a^{12}=a^{36}\)
\(B,5^6:5^3+3^3.3^2=5^3+3^5=125+243=368\)
Tìm X
\(A,\left(x-1\right)^3=125=5^3\)
\(x-1=5\)
\(\Rightarrow x=6\)
\(B,720:\left[41-\left(2x-5\right)\right]=2^3.5=40\)
\(\Leftrightarrow41-\left(2x-5\right)=\frac{720}{40}=18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow x=\frac{28}{2}=14\)
a/\(\frac{1}{9}.3^4.3^n=3^7\)
\(\frac{1}{9}.81.3^n=3^7\)
\(\frac{81}{9}.3^n=3^7\)
\(9.3^n=3^7\)
\(3^2.3^n=3^7\)
\(3^2.3^n=3^2.3^5\)
vậy \(n=5\)
b/ \(\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(2^n\left(\frac{1}{2}+4\right)=9.2^5\)
\(2^n.\frac{9}{2}=9.32\)
\(2^n.\frac{9}{2}=288\)
\(2^n=288:\frac{9}{2}\)
\(2^n=64\)
\(2^n=2^6\)
Vậy \(n=6\)
Chơi câu khó nhất
D = 4 + 42 + 43 + ... + 4n
4D = 42 + 43 + ... + 4n+1
3D = 4n+1 - 4
D = \(\frac{4^{n+1}-4}{3}\)
A=1+2+22+......+2100
=>2A=2+2223+......+2100+2101
=>2A-A=(2+22+23+....+2101)-(1+2+22+.....+2100)
=>A=2101-1
B=3+32+...+350
2B=32+33+..+351
2B-B=(32+33+......+351)-(3+32+...+350)
B=351-3
2B = 3.(3+3^2+...+3^50)
2B = 3+3^2+3^3+...+3^51
2B-B= (3+3^2+3^3+...+3^51) - (3+3^2+3^3+...+3^50)
=> B = 2^51-2
sửa lại giúp mình là
3^51-3 nha
2B = 3.(3+3^2+...+3^50)
2B = 3+3^2+3^3+...+3^51
2B-B= (3+3^2+3^3+...+3^51) - (3+3^2+3^3+...+3^50)
=> B = 2^51-2
\(A=1+2+2^2+2^3+...+2^{100}\)
\(2A=2+2^2+2^3+2^4+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)
\(2A-A=2^{101}-1\)
\(\Rightarrow A=2^{101}-1\)
\(B=3+3^2+3^3+3^4+...+3^{50}\)
\(3B=3^2+3^3+3^4+3^5+...+3^{51}\)
\(3B-B=\left(3^2+3^3+3^4+3^5+...+3^{51}\right)-\left(3+3^2+3^3+3^4+...+3^{50}\right)\)
\(3B-B=3^{51}-3\)
\(\Rightarrow B=3^{51}-3\)
\(C=1+4+4^2+4^3+...+4^n\)
\(4C=4+4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4+4^2+4^3+4^4+...+4^{n+1}\right)-\left(1+4+4^2+4^3+...+4^n\right)\)
\(4C-C=4^{n+1}-1\)
\(C=4^{n+1}-1\)
\(Q=1+2+2^2+2^3+...+2^{50}\)
\(2Q=2+2^2+2^3+2^4+...+2^{51}\)
\(2Q-Q=\left(2+2^2+2^3+2^4+...+2^{51}\right)-\left(1+2+2^2+2^3+...+2^{50}\right)\)
\(\Rightarrow Q=2^{51}-1\)
\(2^n=(2^{51}-1)+1\)
\(\Leftrightarrow2^n=2^{51}\)
Vậy \(n=51\)
~Học tốt~