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Ta có:A= \(1+2+2^2+2^3+...+2^{2010}\)
=> 2A= 2(\(1+2+2^2+2^3+...+2^{2010}\))
=> 2A= 2 +\(2^2+2^3+2^4+...+2^{2011}\)
=> 2A-A= A =(2+ \(2^2+2^3+2^4+...+2^{2011}\)) -( \(1+2+2^2+2^3+...+2^{2010}\))
=> A= \(2^{2011}-1\)
Mà B = \(2^{2011}\)
=> A < B
A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^2010 hay A = 3 + 2^2 + 2^3 + 2^4 + ... + 2^2010 bạn
a) \(\dfrac{7}{5}x\div\left(9-6\dfrac{13}{21}\right)=2\dfrac{13}{25}\)
\(\dfrac{7}{5}x\div\left(9-\dfrac{139}{21}\right)=\dfrac{63}{25}\)
\(\dfrac{7}{5}x\div\dfrac{50}{21}=\dfrac{63}{25}\)
\(\dfrac{7}{5}x=6\)
\(x=\dfrac{30}{7}\)
b) \(\left(1,16-x\right)\times5,25\div\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right)\times2\dfrac{2}{17}=75\%\)
\(\left(1,16-x\right)\times5,25\div\left(\dfrac{95}{9}-\dfrac{29}{4}\right)\times\dfrac{36}{17}=\dfrac{75}{100}\)
\(\left(1,16-x\right)\times5,25\div\dfrac{119}{36}=\dfrac{17}{48}\)
\(\left(1,16-x\right)\times5,25=\dfrac{2023}{1728}\)
\(1,16-x=\dfrac{289}{1296}\)
\(x=0,9370061728\)
52 .(y : 78) = 3380
y : 78 = 3380 : 52
y : 78 = 65
y = 65 . 78
y = 5070
\(\left\{{}\begin{matrix}52⋮4\\52⋮13\\52⋮26\end{matrix}\right.\)
Vì vậy,nếu A chia hết cho 4;13;26;52 thì sẽ chia hết cho 52
\(A=3+3^2+3^3+.....+3^{60}\)
\(A=\left(3+3^2+3^3+3^4+3^5+3^6\right)+\left(3^7+3^8+3^9+3^{10}+3^{11}+3^{12}\right)+...+\left(3^{55}+3^{56}+3^{57}+3^{58}+3^{59}+3^{60}\right)\)\(A=1\left(3+3^2+3^3+3^4+3^5+3^6\right)+3^6\left(3+3^2+3^3+3^4+3^5+3^6\right)+...+3^{54}\left(3+3^2+3^3+3^4+3^5+3^6\right)\)\(A=\left(1+3^6+.....+3^{54}\right)\left(3+3^2+3^3+3^4+3^5+3^6\right)\)
\(A=\left(1+3^6+.....+3^{54}\right).1092\)
\(A=\left(1+3^6+.....+3^{54}\right).21.52\)
\(A⋮4;13;26;52\rightarrowđpcm\)
Gọi số học sinh khối 6 của trường đó là A, ta có:
\(A⋮10\\ A⋮12\\ A⋮15\\ \Rightarrow A⋮BCNN\left(10;12;15\right)\\ \Rightarrow A⋮60\\ \Rightarrow A\in\left\{60;120;180;240;300;360;...\right\}\)
Do \(250\le A\le350\Rightarrow A=300\)
Vậy...
Tính phải k nhỉ?
`1)`
`2x + 3x + 5x`
`= (2 + 3 + 5)x`
`= 10x`
`2)`
`2.x - x + 3.x`
`= (2 - 1 + 3)x`
`= 4x`
`3)`
`9.x - 3 - 3.x`
`= (9 - 3)x - 3`
`= 6x - 3`
`4)`
Thiếu dấu, bạn bổ sung thêm
`5)`
`x - 0,2x - 0,1x`
`= (1 - 0,2 - 0,1)x`
`=0,7x`
`6)`
\(\dfrac{7}{2}x-\dfrac{1}{2}x=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)x=3x\)
xin mọi người đó