(x + 1) + (x + 2) + (x + 3) + ...... + (x + 9) = 90

x + 1 + x + 2 + x + 3 + .... + x + 9 = 90

(x + x + x + .... + x) + (1 + 2 + 3 + ..... + 9) = 90

9x + (1 + 2 + 3 + ..... + 9) = 90

Áp dụng công thức tính dãy số ta có :

\(\frac{\left[\left(9-1\right):1+1\right].\left(9+1\right)}{2}=\frac{9.10}{2}=45\)

=> 9x + 45 = 90

=> 9x = 90 - 45 = 45

=> 9x = 45

=> x = 45 : 9 = 5

`(2/3x-1/2)*3/4-2/5x=4 1/4`

`=>2/3x*3/4-1/2*3/4-2/5x=17/4`

`=>1/2x-3/8-2/5x=17/4`

`=>1/2x-2/5x=17/4+3/8`

`=>x(1/2-2/5)=37/8`

`=>x*1/10=37/8`

`=>x=37/8:1/10`

`=>x=37/8xx10`

`=>x=185/4`

5/8 x 6 + 1/3 = 15/4 + 1/3 = 49/12

5/4- 1/3 = 15/12 - 4/12 = 11/12

a) = 15/4 + 1/3 = 49/12

b) = 5/4 - 1/3 = 11/12

\(\frac{x-12}{3}=\frac{x+1}{4}\)

=>(x-12).4=(x+1)*3

    4x-48=3x+3

    4x-3x=48+3

    x=51

(x-12)/3=(x+1)/4

(x-12)*4=(x+1)*3

x*4-12*4=x*3+1*3

4x-48=3x+3

4x-3x=3+48

x=51

-=> 4 .(x\(-\)12) = 3 .(x+1)

=.>4x\(-\)48      = 3x+3

=>4x\(-\)3x       = 48 + 3

=.> x                  = 51

Vậy x =51

\(\left(x+\frac{1}{3}\right)\left(\frac{3}{4}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{3}=0\\\frac{3}{4}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{3}{8}\end{cases}}\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{3}{8}\right\}\)

Làm đầy đủ hộ mình đi được k bạn

\(a,1\frac{1}{3}+\frac{2}{3}x=5\frac{1}{3}\)

\(\Rightarrow\frac{4}{3}+\frac{2}{3}x=\frac{16}{3}\)

\(\Rightarrow\frac{2}{3}x=\frac{16}{3}-\frac{4}{3}\)

\(\Rightarrow\frac{2}{3}x=4\)

\(\Rightarrow x=4:\frac{2}{3}=4.\frac{3}{2}=6\)

Vậy x=6

\(b,\left(x+\frac{1}{3}\right)\left(\frac{3}{4}-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{3}=0\\\frac{3}{4}-2x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\-2x=\frac{3}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{-3}{8}\end{cases}}\)

Vậy.................

\(c,\frac{-4}{5}x+1\frac{1}{3}x=-1\frac{1}{3}\)

\(\Rightarrow\frac{-4}{5}x+\frac{4}{3}x=\frac{-4}{3}\)

\(\Rightarrow\frac{-2}{3}x=\frac{-4}{3}\)

\(\Rightarrow x=\frac{-4}{3}:\frac{-2}{3}=\frac{-4}{3}.\frac{3}{-2}=2\)

Vậy x=2

a) \(A=4+4^2+4^3+...+4^{200}\)

\(4A=4^2+4^3+...+4^{201}\)

\(4A-A=3A=4^{201}-4\)

\(A=\frac{4^{201}-4}{3}\)

b) \(B=1+5+5^2+...+5^{2017}\)

\(5B=5+5^2+5^3+...+5^{2018}\)

\(5B-B=4B=5^{2018}-1\)

\(B=\frac{5^{2018}-1}{4}\)

c) \(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{500}}\)

\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{499}}\)

\(3C-C=2C=1-\frac{1}{3^{500}}=\frac{3^{500}-1}{3^{500}}\)

\(C=\frac{\left(\frac{3^{500}-1}{3^{500}}\right)}{2}\)

T_i_c_k cho mình nha,có j ko hiểu cứ hỏi mình nhé ^^

\(\left|x+1\right|,\left|x-2\right|,\left|x+3\right|\ge0\)

\(6\ge0\Rightarrow x\ge0\)

\(\left|x+1\right|+\left|x-2\right|+\left|x+3\right|=6\)

\(\Rightarrow\left(x+1\right)+\left(x-2\right)+\left(x+3\right)=6\)

\(\Rightarrow\left(x+x+x\right)+\left(1-2+3\right)=6\)

\(\Rightarrow3x+2=6\)

\(\Rightarrow3x=6-2\)

\(\Rightarrow3x=4\)

\(\Rightarrow x=\frac{4}{3}\)

x+6=xy-y 
=>xy-y-x=6 
=>x(y-1)-(y-1)=7 
=>(x-1)(y-1)=7 
x khác 1 <=> x-1 khác 0,chia 2 vế cho x-1 
Đề bài <=> y = (x+6):(x-1) =1 +7/(x-1) 
x,y nguyên <=> 7 chia hết cho (x-1) <=> (x-1)= cộng,trừ 1 HOẶC (x-1) = cộng,trừ 7 
Giải x,tính y ta có 4 cặp nghiệm: (0;-6) , (2;8) , (8;2) , (-6;0) 
do x,y thuộc N nên chọn (2;8) và (8;2)

mik nha