(x + 1) + (x + 2) + (x + 3) + ... + (x + 26) = 377

=> x + 1 + x + 2 + x + 3 + ... + x + 26 = 377

=> (x + x + x+ ... + x) + (1 + 2 + 3 + ... + 26) = 377

=> 26x + \(\frac{\left(26+1\right).26}{2}\)=377

=> 26x + 351 = 377

=> 26x = 377 - 351

=> 26x = 26

=> x = 1

<  = > 26x + 26.27:2 = 377

26x + 351 = 377

26x = 377 - 351 = 26

x = 26 : 26 = 1 

2x(x-3) - x + 3 = 0

=>2x(x-3)-(x-3)=0

=>(x-3)(2x-1)=0

=>x-3 =0 hoặc 2x-1=0

=>x=3 hoặc x=

vậy S={3;}

`!`

`2x(x-3)-x+3=0`

`=> 2x(x-3)-(x-3)=0`

`=>(x-3)(2x-1)=0`

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

1 < x/5 < 8/5

=> 5/5 < x/5 < 8/5

=> 5 < x < 8

=> x = 6 hoặc 7

1 < 10/x < 2

=> 10/10 < 10/x < 10/5

=> 10 > x > 5

=> x = 6 ; 7 ; 8 hoặc 9

Ủng hộ mk nha ☆_☆★_★^_-

- x có thể bằng 6 hoặc 7

- x nằm trong khoảng từ 1,2,3,4

Các bn giúp tớ với

Ta có công thức:

\(1^3+2^3+.....+n^3=\left(1+2+3+....+n\right)^2\)

\(\Rightarrow1^3+2^3+3^3+.....+10^3=\left(1+2+3+....+10\right)^2=\left(x+1\right)^2\)

\(\Rightarrow x+1=55\Rightarrow x=54\)

\(a,x.\left(x-1\right)=0\)

\(Th1:x=0\)             \(Th2:x-1=0=>x=1\)

Vậy \(x\in\left\{0;1\right\}\)

\(b,\left(x-7\right).\left(x-3\right)=0\)

\(Th1:x-7=0=>x=7\)

\(x-3=0=>x=3\)

Vậy \(x\in\left\{3;7\right\}\)

\(=>2^{x-1}-1=24-9\)
\(2^{x-1}-1=15\)
\(2^{x-1}=16\)
\(=>x-1=4\)
\(x=5\)

\(2^{x-1}-1=24-\left[3^2-\left(2021^0-1\right)\right]\\ 2^{x-1}-1=24-\left[9-\left(1-1\right)\right]\\ 2^{x-1}-1=24-\left[9-0\right]\\ 2^{x-1}-1=24-9\\ 2^{x-1}-1=15\\ 2^{x-1}=15+1\\ 2^{x-1}=16\\ 2^{x-1}=2^4\\ x-1=4\\ x=4+1\\ x=5\)

`2^(x-1) -1 = 24 - [3^2 - (2021^0 -1)]`

`=> 2^(x-1) -1 = 24 - [ 9 - (1-1)]`

`=> 2^(x-1) -1 = 24 - 9`

`=> 2^(x-1) -1 = 15`

`=> 2^(x-1) =15+1`

`=> 2^(x-1) = 16`

`=> 2^(x-1) = 2^4`

`=> x-1=4`

`=> x=4+1`

`=> x=5`

8(x – 3) = 0

x – 3 = 0

x = 3.