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.   \(\frac{x+1}{2}=\frac{8}{x+1}\)

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

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

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

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

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

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

a/

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

$(x+x+....+x)+(1+2+....+100)=5750$
Số lần xuất hiện của $x$:

$(100-1):1+1=100$

Suy ra:

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

$100x+100.101:2=5750$

$100x+5050=5750$

$100x=700$

$x=700:100$

$x=7$

b/

$x^2y-x+xy=6$

$x(xy-1+y)=6$

Do $x,y$ nguyên nên $xy-1+y$ cũng là số nguyên. Mà tích $x(xy-1+y)=6$ nên ta có các TH sau:

TH1: $x=1, xy-1+y=6$

$\Rightarrow y-1+y=6\Rightarrow y=\frac{7}{2}$ (loại) 

TH2: $x=-1, xy-1+y=-6$

$\Rightarrow -y-1+y=-6\Rightarrow -1=-6$ (vô lý - loại) 

TH3: $x=2, xy-1+y=3$

$\Rightarrow 2y-1+y=3\Rightarrow 3y=4\Rightarrow y=\frac{4}{3}$ (loại) 

TH4: $x=-2, xy-1+y=-3$

$\Rightarrow -2y-1+y=-3$

$\Rightarrow -y-1=-3\Rightarrow y=2$ (tm) 

TH5: $x=3, xy-1+y=2\Rightarrow 3y-1+y=2$

$\Rightarrow 4y=3\Rightarrow y=\frac{3}{4}$ (loại) 

TH6: $x=-3, xy-1+y=-2\Rightarrow -3y-1+y=-2$

$\Rightarrow -2y=-1\Rightarrow y=\frac{1}{2}$ (loại) 

TH7: $x=6, xy-1+y=1$

$\Rightarrow 6y-1+y=1\Rightarrow 7y=2\Rightarrow y=\frac{2}{7}$ (loại) 

TH8: $x=-6, xy-1+y=-1$

$\Rightarrow -6y-1+y=-1$

$\Rightarrow -5y=0\Rightarrow y=0$ (tm)

\(a)\) \(A=4+2^2+2^3+...+2^{20}\)

\(A=2^2+2^2+2^3+...+2^{20}\)

\(2A=2^3+2^3+2^4+...+2^{21}\)

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

\(A=2^3+2^{21}-2^2-2^2\)

\(A=2^3+2^{21}-2.2^2\)

\(A=2^3+2^{21}-2^3\)

\(A=2^{21}\)

Vậy \(A=2^{21}\)

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

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

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

\(\Leftrightarrow\)\(100x+5050=5750\)

\(\Leftrightarrow\)\(100x=5750-5050\)

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

\(\Leftrightarrow\)\(x=\frac{700}{100}\)

\(\Leftrightarrow\)\(x=7\)

Vậy \(x=7\)

Chúc bạn học tốt ~ 

b,  (x+x+x+....+x)+(1+2+3+4+...+100)=5750

    100x+5050=5750

    100x=5750-5050

    100x=700

     x=700/100

     x=7

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

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

x . 100 + 5050 = 5750

x . 100            = 5750 - 5050

x .  100           = 700

x                    =  700 : 100

x                      = 7

b) vô câu hỏi tương tự ấy, lười ghi quá :)))

a) x=7

b) 63;66;81

tick cái bạn

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 !

Đặt A=1+2+2²+2³+…+2²⁰

=>2.A=2+2²+2³+2⁴+…+2²¹

=>2.A-A=2+2²+2³+2⁴+…+2²¹-1-2-2²-2³-…-2²⁰

=>A=2²¹-1

Vậy 1+2+2²+2³+…+2²⁰=2²¹-1

(x+1)+(x+2)+(x+3)+…+(x+100)=5750

=>x+1+x+2+x+3+…+x+100=5750

=>(x+x+x+…+x)+(1+2+3+…+100)=5750

Từ 1 đến 100 có:(100-1):1+1=100(số)

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

=>100.x+101.50=5750

=>100.x+5050=5750

=>100.x=5750-5050

=>100.x=700

=>x=7

Vậy x=7

S=3⁰+3²+3⁴+3⁶+...+3²⁰⁰

6S=3²+3⁴+3⁶+...+3²⁰²

6S-S=(3²+3⁴+3⁶+...+3²⁰²)-(1+3²+3⁴+...+3²⁰⁰)

5S=1+(3²-3²)+(3⁴-3⁴)+...+(3²⁰⁰-3²⁰⁰)+3²⁰²

S=(3²⁰⁰+1):5\(\frac{ }{ }\)

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

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

\(\Rightarrow100x+1+2+3+....+100=5750\)

\(\Rightarrow100x+\left[\left(\dfrac{100-1}{1}+1\right):2\right]\left(100+1\right)=5750\)

\(\Rightarrow100x+5050=5750\)

\(\Rightarrow100x=700\)

\(\Rightarrow x=7\)

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

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

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(x=7\)

Vậy ...