tìm x,y hả em

6 đội,16 người

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

\(A=\frac{2^{100}-1}{2^{100}}\)

Tham khảo nhé~

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)

\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow\)\(A=1-\frac{1}{2^{100}}\)

a: \(125\left(19-135\right)-135\left(19-125\right)\)

\(=125\cdot19-125\cdot135-135\cdot19+135\cdot125\)

\(=125\cdot19-135\cdot19\)

\(=19\left(125-135\right)=-10\cdot19=-190\)

b: \(\left(-125\right)\left(135-19\right)-135\left(19-135\right)\)

\(=-125\cdot135+125\cdot19-135\cdot19+135^2\)

\(=135^2-135\cdot125+125\cdot19-135\cdot19\)

\(=135\left(135-125\right)+19\left(125-135\right)\)

\(=135\cdot10-19\cdot10\)

\(=116\cdot10=1160\)

c: \(125\left(19-135\right)+135\left(125-19\right)\)

\(=125\cdot19-125\cdot135+135\cdot125-135\cdot19\)

\(=125\cdot19-135\cdot19\)

\(=19\left(125-135\right)=19\cdot\left(-10\right)=-190\)

d: \(146\left(46-259\right)+259\left(146-46\right)\)

\(=146\cdot46-146\cdot259+259\cdot146-259\cdot46\)

\(=146\cdot46-259\cdot46\)

\(=46\left(146-259\right)\)

\(=46\cdot\left(-113\right)=-5198\)

e: \(195\left(17-185\right)-185\left(17-185\right)\)

\(=195\cdot17-195\cdot185-185\cdot17+185^2\)

\(=\left(195\cdot17-185\cdot17\right)-\left(195\cdot185-185^2\right)\)

\(=17\left(195-185\right)-185\left(195-185\right)\)

\(=10\left(17-185\right)=-168\cdot10=-1680\)

g: \(\left(-146\right)\left(19-156\right)+156\left(19-146\right)\)

\(=-146\cdot19+146\cdot156+156\cdot19-156\cdot146\)

\(=156\cdot19-146\cdot19=19\left(156-146\right)=190\)

h: \(\left(-243\right)\left(25-763\right)+763\left(25-243\right)\)

\(=-243\cdot25+243\cdot763+763\cdot25-763\cdot243\)

\(=763\cdot25-243\cdot25\)

\(=25\left(763-243\right)=520\cdot25=13000\)

Để 1x5y chia hết cho 2 thì y = 0 ,  2 , 4 , 6 , 8

Để 1x5y chia hết cho 5 thì y = 0 , 5 

=> y = 0 

Để 1x5y chia hết cho 3 thì 1 + x + 5 + 0 = 6+ x chia hết cho 3

=> x = 0 , 3 ,6 ,9 

Để 1x5y chia hết cho 6 thì 1 + x + 5 + 0 = 6+x chia hết cho 6 

=> x = 0 ; 6 

Để 1x5y chia hết cho 9 thì 1 + x + 5 + 0 = 6 + x chia hết cho 9 

=> x = 3 

=> Ko tồn tại x 

Tham khảo của cj Nguyễn :33

(x-3)^11=(x-3)^7

(x-3)^11-(x-3)^7=0

(x-3)^7[(x-3)^4-1)]=0

\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^7=0\\\left(x-3\right)^4-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^4=1\end{cases}}\)\(\Rightarrow\)x=3; x=2; x=4

Vậy x=3 hoặc x=2 hoặc x=4

Ta có (x-3)^11 = (x-3)^7

  <=>  \(\hept{\begin{cases}x-3=0\\x-3=1\\x-3=-1\end{cases}}\)

   <=> \(\hept{\begin{cases}x=3\\x=4\\x=2\end{cases}}\)

a: =5^3(5^3+1)=125*126=15750

b: =10^3(10^2-10+1)=1000*91=91000

c: =15^n(15^2+1)=226*15^n

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