Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
Nếu là z+x thì mik biết làm nè:
Đặt x-y=2011(1)
y-z=-2012(2)
z+x=2013(3)
Cộng (1);(2);(3) lại với nhau ta được :
2x=2012=>x=1006
Từ (1) => y=-1005
Từ (3) => z=1007
a)\(123-5:\left(x+4\right)=38\)
\(5:\left(x+4\right)=123-38\)
\(5:\left(x+4\right)=85\)
\(x+4=5:85\)
\(x=\dfrac{1}{17}-4\)
\(x=-\dfrac{67}{17}\)
b)\(70-5.\left(x-3\right)=45\)
\(5.\left(x-3\right)=70-45\)
\(5.\left(x-3\right)=35\)
\(x-3=35:5\)
\(x-3=7\)
\(x=7+3\)
\(x=10\)
Tuy có vẻ hơi muộn nhưng thôi ![]()
Nếu A là số tự nhiên ⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
\(\Rightarrow7^{2004}-3^{92^{94}}⋮10\)
Thật vậy, ta có :
72004 với lũy thừa là 2004 ⋮ 4
⇒ 72004 = ( .......... 9 )
392^94 với lũy thừa là 9294 mà 92 ⋮ 4 ⇒ 9294 ⋮ 4
⇒ 392^94 = ( .......... 9 )
⇒ 72004 - 392^94 = ( .......... 9 ) - ( ............ 9) = ( ........... 0 ) ⋮ 10
⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
A=1/10.(72004-392^94) là số tự nhiên.



giải giùm tớ nha

Bài 4:
a: \(A=2+2^2+2^3+\cdots+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+\cdots+2^{59}\right)\) ⋮3
Ta có: \(A=2+2^2+2^3+\cdots+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+\cdots+2^{58}\right)\) ⋮7
Ta có: \(A=2+2^2+2^3+\cdots+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+\cdots+2^{57}\right)\) ⋮15
b: \(B=3+3^3+3^5+\ldots+3^{1991}\)
\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+\cdots+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+\cdots+3^{1986}\left(3+3^3+3^5\right)\)
\(=273\left(1+3^6+\cdots+3^{1986}\right)\) ⋮13
\(B=3+3^3+3^5+\ldots+3^{1991}\)
\(=\left(3+3^3+3^5+3^7\right)+\left(3^9+3^{11}+3^{13}+3^{15}\right)+\cdots+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(=\left(3+3^3+3^5+3^7\right)+3^8\left(3+3^3+3^5+3^7\right)+\ldots+3^{1984}\left(3+3^3+3^5+3^7\right)\)
\(=\left(3+3^3+3^5+3^7\right)\left(1+3^8+\cdots+3^{1984}\right)\)
\(=2460\left(1+3^8+\cdots+3^{1984}\right)\) ⋮41