\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)

\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)

\(100x-5050=50\)

\(100x=50+5050\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)

\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=7\)

\(b,x+\left(1+2+3+...+50\right)=2000\)

 \(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)

\(x+1275=2000\)

\(\Rightarrow x=2000-1275=725\)

c , ( x - 1 ) + ( x - 2 ) + ...... + ( x - 100 ) = 50

( x + x + ........ + ) - ( 1 + 2 + ..... + 100 ) = 50

100x - 5050 = 50

100x = 50 + 5050

100x = 5100

=> x = 5100/100 = 51

Ta có:

1/2+1/4+1/8+1/16+1/32=31/32

Vì có 5 tổng=> ta gọi phần còn lại là 5X

=>5X+31/32=2

=>5X+31/32=64/32

=>5X=64-32-31/32

=>5X=33/32

=>X=33/32:5

=>X=33/160

  64/32 chứ không phải là 64-32 đâu nha

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

a) 2^x . 16^2 = 1024                          b) 64 . 4^x =  16^8                    c) 2^x = 16

=> 2^x . 256 = 1024                             => 64 . 4^x  = (4^2) ^ 8               => 2^x  = 2^4

=> 2^x          = 1024 : 256                    =>   4^3 . 4^x  =  4^16                  => x     = 4

=> 2^x           =  4                                 =>             4^x  = 4^16 : 4^3

=>  2^x           = 2^2                             =>              4^x  =  4^13

                                                              =>                 x   =  13 

=>      x           = 2

a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)

b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)

c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)

`a)  2^x div 4=16`

`<=> 2^x=64`

`<=> 2^x=2^6`

`<=> x=6`

`b)  |x+1|=-2`

do `|x+1|>=0 AA x in ZZ`

$\to x\in\varnothing$

`c)  2x+15=-27`

`<=> 2x=-42`

`<=> x=-21`

1 là đăng lại ảnh 

2 là tách nhỏ ra

II . Hình học 

Bài 2 :

BC = AB - AC = 10 - 8 = 2 (cm )

Vì C là trung điểm của BN

nên CN = CB = 2 (cm)

NB = NC + BC = 2+ 2 = 4( cm )

Vậy CN = 2m và NB = 4 cm

có cái khác rồi

Bài 1:

a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)

b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)

Bài 2:

a)\(2^{x-1}=16\)

\(\Rightarrow2^{x-1}=2^4\)

\(\Rightarrow x-1=4\Rightarrow x=5\)

b)\(\left(x-1\right)^2=25\)

\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)

\(\Rightarrow x=6\) hoặc \(x=-4\)

Vậy \(x=6\) hoặc \(x=-4\)

c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)

d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)

Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)

Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)

\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)