a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555

=> 101x +5050 = 5555

=> 101x = 505

=> x = 505 : 101 = 5

Vậy, x = 5

b)1+2+3+4+...+x=820

=> ( x+1) x :2 = 820

=> (x+1)x = 1640

Mà 1640 = 40 . 41

=> x = 40 ( vì {x+1} - x = 1)

Vậy, x = 40

c) 3^x+1 = 9.27=243

=> 3^x+1 = 3⁵

=>x + 1 = 5

=> x = 4

Vậy, x=4

d) x+2x+3x+...+99x+100x=15150

=> [( 100 + 1) x 100 :2 ] x = 15150

=> 5050x = 15150

=> x = 15150:5050 = 3

Vậy, x =3

e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550

=> 100x + 5050 = 205550

=> 100x =  205550 - 5050= 200500

=> x =  200500 : 100 = 2005

Vậy, x = 2005

f)3^x+3^x+1+3^x+2=351

=> 3^x + 3^x . 3 + 3^x x 9 = 351

=> 3^x ( 1+3+9) = 351

=> 3^x  . 13 = 351

=> 3^x  = 351 :13=27 mà 27 = 3³

=> x=3

Vậy, x=3

mình đg cần gấp á

a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)

\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)

\(\Rightarrow101\cdot x+5050=5555\)

\(\Rightarrow101\cdot x=5555-5050\)

\(\Rightarrow101\cdot x=505\)

\(\Rightarrow x=505:101\)

\(\Rightarrow x=5\)

b) \(1+2+3+4+...+x=820\)

\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)

\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)

\(\Rightarrow\left(x+1\right)\cdot x:2=820\)

\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=1640\)

Ta thấy: \(40\cdot41=1640\)

Vậy: \(x=40\)

a) 154 - 452 + ( -198 ) - ( -190 )

=  154 - 452 + ( - 198) + 190

=    ( - 298) +  (-8)

=            (- 316)

b) - 351 - [ ( -98) + 75] - ( -176)

=  - 351 - [ (-23)] + 176

=  - 351 + 23 + 176

=   - 328 + 176

=   -   152

c) 77 + ( - 171 ) - 96 - 757

=  ( - 94 ) + ( - 96 ) + ( - 757 )

=  ( - 190 ) + ( - 757 )

=     ( - 947 )

d) - 512 - [ 198 - 275 + (-190) - (-541) ]

= - 512 -  [ (-77) + (-190) + 541 ]

=  - 512 -  [ (-267) + 541 ]

= - 512 - 274

= - 521 + ( - 274 )

=  - 795

a)  4 1 5 + 3 2 5 = 21 5 + 17 5 = 38 5

b)  1 2 − 1 8 − 1 16 = 8 16 − 2 16 − 1 16 = 5 16

c)  426 × 305 = 129930

d)  72306 : 351 = 206

1) \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x\left(1+3^1+3^2\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

2) \(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(\Rightarrow C=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)...+2^{96}\left(2+2^2+2^3+2^4\right)\)

\(\Rightarrow C=30+2^4.30...+2^{96}.30\)

\(\Rightarrow C=\left(1+2^4+...+2^{96}\right).30⋮30\)

mà \(30=5.6\)

\(\Rightarrow C⋮5\left(dpcm\right)\)

1,

Có \(3^x\)+ \(3^{x+1}\) + \(3^{x+2}\) = \(351\)

=> \(3^x\) + \(3^x\).\(3\) + \(3^x\).\(9\) = \(351\)

=> \(3^x\).\(13\) = \(351\)

=> \(3^x\) = \(27\)

=> \(x\) = \(3\)

2,

C = \(2\) + \(2^2\) + \(2^3\) + ... + \(2^{100}\)

2C = \(2^2\) + \(2^3\) + \(2^4\) + ... + \(2^{101}\)

2C - C = \(2^{101}\) - \(2\)

C = \(2^{101}\) - \(2\)

C = \(2\).\(\left(2^{100}-1\right)\)

C = 2.\(\left(\left(2^5\right)^{20}-1^{20}\right)\)

Có \(2^5\) \(-1\) \(⋮\) 5

=> \(\left(\left(2^5\right)^{20}-1^{20}\right)\) \(⋮\) 5

=> C \(⋮\) 5

3,

Xét \(\overline{abcdeg}\)

= \(\overline{ab}\).\(10000\) + \(\overline{cd}\).\(100\) + \(\overline{eg}\)

= \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\) + \(9.\left(1111.\overline{ab}+11.\overline{cd}\right)\)

Có\(\left\{{}\begin{matrix}9.\left(1111.\overline{ab}+11.\overline{cd}\right)⋮9\left(1111.\overline{ab}+11.\overline{cd}\inℕ^∗\right)\\\overline{ab}+\overline{cd}+\overline{eg}⋮9\end{matrix}\right.\)

=> \(\overline{abcdeg}⋮9\)

4,

S = \(3^0+3^2+3^4+...+3^{2002}\)

9S = \(3^2+3^4+3^6+...+3^{2004}\)

9S - S = \(3^2+3^4+3^6+...+3^{2004}\) - (\(3^0+3^2+3^4+...+3^{2002}\))

8S = \(3^{2004}-1\)

=> 8S \(< 3^{2004}\)

a) 2011 + 5 . [300- ( 18- 8)2]

= 2011 + 5. ( 300 - 102)

= 2011 + 5. 200

= 3011

b) Số số hạng trong tổng trên là:

( 99 - 1) ; 2 = 1 = 50 (số)

99 + 97 = … = 3 + 1

= ( 99 + 1) . 50 : 2

= 2500

c) Số số hạng trong tổngtrên là:

( 100 - 1) : 3 + 1 = 34 ( số)

100 + 97 + 94 + …+ 4 = 1

= ( 100 + 1) . 34 : 2=1717

d) 99 - 97 + 95 - 93 + … + 3 - 1

= 2 + 2 + 2 + … + 2

= 2. 25

= 50

e) 100 - 97 + 94 - …+ 4 - 1

= 3 + 3 + 3 + … + 3 

= 3. 17 

= 51

1)

\(\left(a\right)37+397+3997+39997\)

\(=40-3+400-3+4000-3+40000-3\)

\(=\left(40+400+4000+40000\right)-\left(3+3+3+3\right)\)

\(=44440-12=44428\)

\(\left(b\right)298+2998+29998+299998\)

\(=300-2+3000-2+30000-2+300000-2\)

\(=\left(300+3000+30000+300000\right)-\left(2+2+2+2\right)\)

\(=333300-8=333296\)

\(\left(c\right)9+99+999+9999+99999\)

\(=10-1+100-1+1000-1+10000-1+100000-1\)

\(=\left(10+100+1000+10000+100000\right)-\left(1+1+1+1+1\right)\)

\(=111110-5=111105\)

2)
\(\left(a\right)\left(2+4+6+...+2002+2004+2006\right)-\left(1+3+5+...+2001+2003+2005\right)\)

\(=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(2002-2001\right)+\left(2004-2003\right)+\left(2006-2005\right)\)

\(=1+1+1+...+1+1+1\)( 1003 số 1 )

\(=1003\)

\(\left(b\right)88-87+86-85+84-83+...+6-5+4-3+2-1\)

\(=\left(88-87\right)+\left(86-85\right)+\left(84-83\right)+...+\left(6-5\right)+\left(4-3\right)+\left(2-1\right)\)

\(=1+1+1+...+1+1+1\)( 44 số 1 )

\(=44\)

\(\left(c\right)100-98+96-94+92-90+...+12-10+8-6+4-2\)

\(=\left(100-98\right)+\left(96-94\right)+\left(92-90\right)+...+\left(12-10\right)+\left(8-6\right)+\left(4-2\right)\)

\(=2+2+2+...+2+2+2\) ( 25 số 2 )

\(=50\)

3)

\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)

\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)

\(=3+3+3+3+3+3+3+3+3+3+3=33\)

\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)

\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)

\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)

\(=2006-1275=731\)

\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)

\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)

\(=4+4+4+4+4+4+4+4+4+4+4=44\)

Cảm ơn bạn nha