Mình chỉ ghj đáp za thôj nên thông cảm nha

b)1953368

c)225

d)32

\(a,=4^{10}.4^{10}.4^{45}\)

    \(=4^{65}\)

\(b,=5^9+3^5\)

    \(=1953125+243\)

     \(=1953368\)

\(c,=1+8+27+64+125\)

    \(=225\)

\(d,=32^5:32^4\)

     \(=32\)

\(\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+....+\left(2^2+2\right)\)

\(=2^9.\left(2+1\right)+2^7.\left(2+1\right)+...+2.\left(2+1\right)\)

\(=2^9.3+2^7.3+...+2.3\)

\(=3.\left(2^9+2^7+...+2\right)⋮3\)

P/S: mấy bài khác tương tự

\(a,2^{10}+2^9+2^8+...+2\)

\(=\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+...+\left(2^2+2\right)\)

\(=2^9\left(2+1\right)+2^7\left(2+1\right)+...+2\left(2+1\right)\)

\(=2^9.3+2^7.3+...+2.3\)

\(=3\left(2^9+2^7+...+2\right)⋮3\left(đpcm\right)\)

\(b,1+3+3^2+3^3+...+3^{99}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)

\(=4+3^2\left(1+3\right)+...+3^{98}\left(1+3\right)\)

\(=4+3^2.4+...+3^{98}.4\)

\(=4\left(1+3^2+...+3^{98}\right)⋮4\left(đpcm\right)\)

\(c,1+5+5^2+5^3+...+5^{1975}\)

\(=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{1974}+5^{1975}\right)\)

\(=6+5^2\left(1+5\right)+...+5^{1974}\left(1+5\right)\)

\(=6+5^2.6+...+5^{1974}.6\)

\(=6\left(1+5^2+...+5^{1974}\right)⋮6\left(đpcm\right)\)

Hình như bạn thiếu số hạng 4 trong tổng A nhé 

\(4A=4+4^3+4^4+...+4^{100}\)

\(\Rightarrow4A-A=\left(4+4^2+4^3+4^4+...+4^{100}\right)-\left(1+4+4^2+4^3+...+4^{99}\right)\)

\(\Rightarrow3A=4^{100}-1\)

\(\Rightarrow A=\frac{4^{100}-1}{3}\)

Mà B = 4¹⁰⁰ nên \(A=\frac{B-1}{3}\Rightarrow A=\frac{B}{3}-\frac{1}{3}\)  do đó \(A< \frac{B}{3}\)

nói là thiêus số hạng 4 mà ở 4A có thấy 4² đâu ?

Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)

\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)

\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)

\(\Rightarrow24A=5^{51}-5\)

\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)

Vậy ............................................................

1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)

\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)

Vậy \(x=3\)

b) \(\left(4x-1\right)^3=-27.125\)

\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)

\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)

Vậy \(x=-3,5\)

c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)

\(\Rightarrow4x+4=4x+12\)

\(\Rightarrow4x=4x+8\)

\(\Rightarrow x\in\varnothing\)

d) \(\left(x-5\right)^7=\left(x-5\right)^9\)

\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)

\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

ta có: M = 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +......+ 99/3^99 - 100/3^100

=> 3.M = 1 - 2/3 + 3/3^2 - 4/3^3 +.......+ 99/3^98 - 100/3^99

=> 3M + M = ( 1 - 2/3 + 3/3^2 - 4/3^3 +.........+ 99/3^98 - 100/3^99 ) + ( 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +....+ 99/3^99 - 100/3^100 )

=> 4.M = 1- 1/3 + 1/3^2 - 1/3^3 +........+ 1/3^98 - 1/3^99 - 100/3^100

=> 12.M = 3 - 1 + 1/3 - 1/3^2 +.......+ 1/3^97 - 1/3^98 - 1/3^99

=> 12M + 4M = ( 3 - 1 + 1/3 - 1/3^2 +......+ 1/3^97 - 1/3^98 - 1/3^99 ) + ( 1 - 1/3 + 1/3^2 - 1/3^3 +.......+ 1/3^99 - 1/3^100 )

=> 16M = 3 - 101/3^99 - 100/3^100

vù 16M < 3

=> M < 3/16

vậy M < 3/16

tk cho mk nha,mk bị âm rùi

Bài 1:

C = 1/101 + 1/102 + 1/103 + ... + 1/200

Có:

C < 1/101 + 1/101 + 1/101 + ... + 1/101

C < 100 . 1/101

C < 100/101

Mà 100/101 < 1

=> C < 1 (1)

Có:

C > 1/200 + 1/200 + 1/200 + ... + 1/200

C > 100 . 1/200

C > 1/2 (2)

Từ (1) và (2)

=> 1/2<C<1

Ủng hộ nha mk làm tiếp

J=6 + 16 + 30 + 48 +...+ 19600 + 19998

Chia cả 2 vế cho 2 ta được

B/2 = 3 + 8 + 15 + 24 +  ......... + 98000+ 9999

B/2= 1x3+2x4+3x5+4x6+…….+98x100+99x101

B/2= 100/6[(100-1)x(2x100+1)] = 328350

-> B =328350x2=656700

K=2 + 5 + 9 + 14 + ....+ 4949 + 5049

Nhân cả 2 vế với 2 ta được

2xD=1x4+    2x5+ 3x6+   4x7+……..+98x101+99x102

2xD = 1(2+2)+2(3+2)+3(4+2)+...+99(100+2)

2xD = 1x2+1x2+2x3+2x2+3x4+3x2+...+99x100+99x2

2xD= (1x2+2x3+3x4+...+99x100)+2(1+2+3+...+99)

2xD =           333300       +                      9900        =      343200

 -> D= 343200 :2 =171600