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a) 5^x=5^78:5^14(lấy 78-14)
5^x=5^64
=> x=64
b) 7^x.7^2=7^21
7^x=7^21:7^2
7^x=7^19
=> x=19
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
Các bn ơi giải hộ mik với. Ai giải đầu mik sẽ k cho. Cảm ơn các bn nhiều nha.
\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{18}}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{131072}-\dfrac{1}{262144}=1+1-\dfrac{1}{262144}=2-\dfrac{1}{262144}\)
Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
3.42+(57:56)-(2.24)
=3.42+57-6-24+1
=3.42+51-25
=(3.42)+5-32
=48+5-32
=53-32
=21
\(A=7^3+7^4+7^5+7^6+...+7^{97}+7^{98}\)
\(=\left(7^3+7^4\right)+\left(7^5+7^6\right)+....+\left(7^{97}+7^{98}\right)\)
\(=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\)
\(=\left(1+7\right)\left(7^3+7^5+...+7^{97}\right)\)
\(=8\left(7^3+7^5+...+7^{97}\right)⋮8\)
Vì A có: 96 số hạng nên ta chia A thành 48 nhóm 1 nhóm có 2 số hạng
\(A=7^3+7^4+7^5+7^6+...........+7^{97}+7^{98}\)
\(A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...........+\left(7^{97}+7^{98}\right)=7^3\left(1+7\right)+7^5\left(1+7\right).....+7^{97}\left(1+7^{ }\right)\)
\(A=7^3.8+7^5.8+.......+7^{97}.8=8\left(7^3+7^5+........+7^{97}\right)⋮8\left(ĐPCM\right)\)
CÂU 1
\(5^{n+1}+5^n=750\)
\(=>5^n\cdot5+5^n=750\)
\(=>5^n\cdot\left(5+1\right)=750\)
\(=>5^n\cdot6=750\)
\(=>5^n=750:6\)
\(=>5^n=125\)
\(=>5^n=5^3\)
\(=>n=3\)
a)\(S=1+5+5^2+...+5^{10}\)
\(5S=5+5^2+5^3+...+5^{11}\)
\(5S-S\)hay 4S\(=5^{11}-1\)
\(\Rightarrow S=\left(5^{11}-1\right):4\)
b)\(S=1+7+7^2+...+7^{10}\)
\(7S=7+7^2+7^3+...+7^{11}\)
\(7S-S\)hay 6S\(=7^{11}-1\)
\(\Rightarrow S=\left(7^{11}-1\right):6\)
Học tốt nha!!!