\(\frac{ }{aaa}⋮37\)

b)

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3 tháng 11 2016

a, Ta có : \(\overline{aaa}=a.111=a.3.37\Rightarrow\overline{aaa}⋮37\)

b,Vì : \(\overline{aaaaaa}=a.111111=a.15873.7\Rightarrow\overline{aaaaaa}⋮7\)

c,Vì : \(\overline{abcabc}=\overline{abc}.1001\Rightarrow\overline{abcabc}⋮1001\)

d, Ta có : \(\overline{ab}+\overline{ba}=10a+b+10b+a\)

\(=10a+a+10b+b=11a+11b\)

\(=11\left(a+b\right)⋮11\) ( Vì : \(a+b\in N\) )

Vậy \(\overline{ab}+\overline{ba}⋮11\)

e, \(\overline{ab}-\overline{ba}=\left(10a+b\right)-\left(10b+a\right)\)

\(=\left(10-1\right)a-\left(10-1\right)b\)

\(=9a-9b=9\left(a-b\right)\)

Vì : \(a\ge b\Rightarrow a-b\in N\Rightarrow9\left(a-b\right)⋮9\)

Vậy : \(\overline{ab}-\overline{ba}⋮9\)

f, \(\overline{abc}-\overline{cba}=\left(a.100+b10+c\right)-\left(100c+10b+a\right)\)

\(=\left(100a+10a+10c+c\right)-\left(100c+10c+10a+a\right)\)

\(=\left(110a+11c\right)-\left(110c+11a\right)⋮11\)

Vì : \(a\ge c\Rightarrow\overline{abc}-\overline{cba}⋮11\)

Vậy : \(\overline{abc}-\overline{cba}⋮11\)

3 tháng 11 2016

a) \(\overline{aaa}=a.111⋮37\)

\(\Rightarrow\overline{aaa}⋮37\left(đpcm\right)\)

b) \(\overline{aaaaaa}=a.111111⋮7\) ( vì \(111111⋮7\) )

\(\Rightarrow\overline{aaaaaa}⋮7\left(đpcm\right)\)

c) \(\overline{abcabc}=\overline{abc}.1001⋮1001\)

\(\Rightarrow\overline{abcabc}⋮1001\left(đpcm\right)\)

d) \(\overline{ab}+\overline{ba}=10a+b+10b+a=11a+11b=11\left(a+b\right)⋮11\)

\(\Rightarrow\overline{ab}+\overline{ba}⋮11\left(đpcm\right)\)

e) \(\overline{ab}-\overline{ba}=10a+b-\left(10b+a\right)=9a-9b=9\left(a-b\right)⋮9\)

\(\Rightarrow\overline{ab}-\overline{ba}⋮9\left(đpcm\right)\)

f) \(\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=11\left(9a-9b\right)⋮11\)

\(\Rightarrow\overline{abc}-\overline{cba}⋮11\left(đpcm\right)\)

30 tháng 7 2016

Theo bài , ta có :

a > b 

và ab + ba = 110

=) b + a = 0 và nhớ 1 

nên =) chỉ có thể là 4 + 6 

mà 6 > 4 

=) a = 6 ; b = 4 

30 tháng 7 2016

bn có chắc chắn hông zậy??????????

27 tháng 6 2016

a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

b)\(\left(2x-3\right)^3=343\)

\(\left(2x-3\right)^3=7^3\)

\(2x-3=7\)

\(2x=7+3\)

\(2x=10\)

\(x=10:2\)

\(x=5\)

27 tháng 6 2016

a) Ta có: \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

<=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

<=> \(x=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy x=5/6

b)\(\left(2x-3\right)^3=343\)

<=>\(2x-3=\sqrt[3]{343}=7\)

<=> 2x=10 <=> x=5

c) \(\left(\frac{1}{3}\right)^{2x}+1=\frac{1}{7}\)

<=>\(\left(\frac{1}{3}\right)^{2x}=\frac{-6}{7}\)

<=> \(\left(\frac{1}{3^x}\right)^2=-\frac{6}{7}\)(vô lí vì \(\left(\frac{1}{3^x}\right)^2\ge0\))

Vậy ko tìm được x thỏa mãn.

d)\(\left(2x-3\right)^2=9\)

=>\(\left[\begin{array}{nghiempt}2x-3=3\\2x-3=-3\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=3\\x=0\end{array}\right.\)

Vậy x=3 hoặc x=0.

e) \(\left(x-3\right)^6=\left(x-3\right)^7\)

<=> \(\left(x-3\right)^7-\left(x-3\right)^6=0\)

<=> \(\left(x-3\right)^6\left(x-3-1\right)=0\)

<=>\(\left(x-3\right)^6\left(x-4\right)=0\)

<=> \(\left[\begin{array}{nghiempt}x-3=0\\x-4=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=4\end{array}\right.\)

Vậy x \(\in\left\{3;4\right\}\)

27 tháng 1 2017

b) Ta có:

\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)

\(\Rightarrow B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)

\(\Rightarrow B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)

\(\Rightarrow B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)

Vậy \(\frac{A}{B}=\frac{1}{2017}\)

10 tháng 6 2016

\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)

Xét \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)

và \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.10=\frac{10}{19}>\frac{1}{2}\)

Do đó \(B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{5}{4}>1\)

5 tháng 5 2016

\(\frac{2x+1}{3}=\frac{5}{2}\)

\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)

2x=  15/2 - 1 = 13/2

x = 13/2 : 2

x = 13/4 

b) 2x + 2x+1 + 2x+2 + 2x+3 = 480

2x.(1+ 2 +22 + 23) = 480

2x . 15 = 480

2x = 480 : 15 = 32

2x = 25 => x = 5

c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)

\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)

\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)

< = > 3x=  -6 => x = -2

 

5 tháng 5 2016

Hỏi đáp Toán

30 tháng 8 2016

2. \(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.\left(1993+1\right)-1}{1993.1995+1994}=\frac{1995.1993+1995-1}{1993.1995+1994}=\frac{1995.1993+1994}{1993.1995+1994}\)

7 tháng 10 2016

1. \(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}\) 

\(=\frac{7-3}{3.7}+\frac{12-7}{7.12}+\frac{13-12}{12.13}+\frac{23-20}{20.23}\) 

\(=\left[\frac{7}{3.7}-\frac{3}{3.7}\right]+\left[\frac{12}{7.12}-\frac{7}{7.12}\right]+\left[\frac{13}{12.13}-\frac{12}{12.13}\right]+\left[\frac{20}{13.20}-\frac{13}{13.20}\right]+\left[\frac{23}{20.23}-\frac{20}{20.23}\right]\) \(=\left[\frac{1}{3}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{12}\right]+\left[\frac{1}{12}-\frac{1}{13}\right]+\left[\frac{1}{13}-\frac{1}{20}\right]+\left[\frac{1}{20}-\frac{1}{23}\right]\) \(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\) \(=\frac{1}{3}-\frac{1}{23}\\ =\frac{20}{69}\)

26 tháng 1 2017

mi chưa mần à?

2 tháng 6 2016

\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)

\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)

2 tháng 6 2016

giúp mình đi@@@@huhu