Đề sai rồi! Đề đúng là : (x+1)+(x+2)+......+(x+100)=5750

=> (x+x+x+.....+x) + (1+2+3+......+100) = 5750

=> 100x + 5050 = 5750

=> 100x = 700

=> x = 7 (tm)

Vậy...................

a, \(2A=8+2^3+2^4+...+2^{21}\)

\(\Rightarrow2A-A=2^{21}+8-\left(4+2^2\right)+\left(2^3-2^3\right)+...+\left(2^{20}-2^{20}\right)=2^{21}\)

b, (x+1)+(x+2)+(x+3)+...+(x+100)=5750 

= 100x + (1+2+...+100) = 5750 

=> x = [ 5750 - (1+2+...+100) ] :100 

=> x = (5750 - 5050) : 100 

=> x = 7

a) (x+1)+(x+2)+....+(x+100)=5750

<=> (x+x+x+....+x)+(1+2+....+100)=5750

<=> 100x+5050=5750

<=> 100x=700

<=> x=7

b) A=7-Ix-1I

Ta có Ix-1I =<0 với mọi x thuộc Z
=> 7-Ix-1I =<7 với mọi x thuộc Z hay A =< 7

Dấu "=" <=> Ix-1I=0

<=> x-1=0

<=> x=1

Vậy MaxA=7 đạt được khi x=1

(𝑥+1)+(𝑥+2)+⋯+(𝑥+100)=5750

=) x.100 + ( 100 + 99 + .... + 2 + 1 ) = 5750

=) x.100 + 5050 = 5750

=) x.100 = 5750 - 5050 = 200

=) x = 200/100 = 2 

Vậy x = 2

Nếu mình sai thì các bạn sẽ cùng góp ý với mình nhoa !

b)Ta có:\(\left|x-1\right|\ge0\)

\(\Rightarrow-\left|x-1\right|\le0\)

\(\Rightarrow7-\left|x-1\right|\le7\)

\(\Rightarrow A\le7\)

Dấu bằng xảy ra khi:\(\left|x-1\right|=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

A=4+2²+2³+2⁴+...+2²⁰

=2²+2²+2³+2⁴+...+2²⁰

=>2A=2³+2³+2⁴+...+2²¹

=>2A-A=2³+2³+2⁴+...+2²¹-2²-2²-2³-2⁴-...-2²⁰

=>A(2-1)=2³+2²¹-2²-2²

=>A=8+2²¹-4-4

=>A=2²¹

câu B làm thiếu bước rồi. Bạn làm tắt quá

b) ( x+1 ) + ( x+2 ) + ...+ ( x+100 ) = 5750

( x+x+x+...+x ) + ( 1+2+3+...+100 ) = 5750

( x.100 ) + (1.100) . 100 : 2 = 5750

( x.100 ) + 5050 = 5750

( x.100 ) = 5750-5050

 x.100 = 700

x = 700 : 100

x=7

Vậy x = 7.

a)  (x+1) +(x+2) +...+(x+100) = 5750

100.x + (1+2+...+100) =5750

100.x + (100.101):2 = 5750

100.x + 5050 = 5750

100x= 700

x= 7

a) Dãy trên có số số hạng là :

( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x 

=> 1 + 2 + 3 + 4 + ... + x = 45

=> ( x + 1 ) . x : 2  = 45

=> ( x + 1 ) . x = 90

=> ( x + 1 ) . x = 10 . 9

=> x = 9

Vậy x = 9

a) Dãy trên có số số hạng là :

( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x 

=> 1 + 2 + 3 + 4 + ... + x = 45

=> ( x + 1 ) . x : 2  = 45

=> ( x + 1 ) . x = 90

=> ( x + 1 ) . x = 10 . 9

=> x = 9

Vậy x = 9

b) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 ( mỗi nhóm có 100 số hạng )

=> 100x + ( 100 + 1 ) . 100 : 2 = 5750

=> 100x + 5050 = 5750

=> 100x = 700

=> x = 7

Vậy x = 7

a)5^x=125

5^x=5^3

=> x=3

b) (x+1)+(x+2)+(x+3)+...+(x+100)=5750

(x+x+x+...+x)+(1+2+3+...+100)=5750

100x+5050=5750

100x=5750-5050

100x=700

x=700:100

x=7

a,2A=8+2³+2⁴+2⁵+............+2²¹

2A-A=(8+2³+2⁴+2⁵+...........+2²¹)-(4+2²+2³+2⁴+.............+2²⁰)

A=(8+2³+2²¹)-(4+2²)

A=8+8+2²¹-4-4

A=4+4+2²¹

A=2³+2²¹

b,(x+1)+(x+2)+.............+(x+100)=5750

=>x+1+x+2+...........+x+100=5750

=>100x+(1+2+............+100)=5750

=>100x+\(\frac{100.\left(100+1\right)}{2}\)=5750

=>100x+5050=5750

=>100x=5750-5050

=>100x=700

=>x=700:100

=>x=7

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

\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+..+100\right)=5750\Rightarrow x.100+\left(100+1\right)\cdot100:2=5750\)\

\(\Rightarrow x.100+5050=5750\Rightarrow x.100=700\Rightarrow x=7\)

b) \(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)

\(\Rightarrow\left(x+1\right)^2=16\Rightarrow\left(x+1\right)^2=4^2\)

\(\Leftrightarrow x+1=4\Rightarrow x=3\)

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

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

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=5750-5050=700\)

\(\Leftrightarrow x=700:100=7\)

2.   \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)

\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)

\(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow\left(x+1\right)=16:2\)

\(\Leftrightarrow\left(x+1\right)=8\)

\(\Leftrightarrow x=8-1=7\)

Bạn sửa lại chỗ bài 2. Ở chỗ:

\(\left(x+1\right)^2=16\) và các chỗ sau chỗ đó thành:

\(\left(x+1\right)^2=16\)

\(\Leftrightarrow\left(x+1\right)^2=4^2\)

\(\Leftrightarrow x+1=4\)

\(\Leftrightarrow x=4-1=3\)