Sửa đề: x=2024

x=2024 nên x+1=2025

Ta có: \(x^6-2025x^5+2025x^4-2025x^3+2025x^2-2025x+2025\)

\(=x^6-x^5\left(x+1\right)+x^4\left(x+1\right)-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

=1

Sửa đề: Cho x,y,z đôi một khác nhau và \(x^3+y^3+z^3=3xyz\)

Ta có: \(x^3+y^3+z^3=3xyz\)

=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

=>\(\left(x+y+z\right)\left\lbrack\left(x+y\right)^2-z\left(x+y\right)+z^2\right\rbrack-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left\lbrack x^2+2xy+y^2-xz-zy+z^2\right\rbrack-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left\lbrack x^2+y^2+z^2-xy-xz-yz\right\rbrack=0\)

=>\(\left(x+y+z\right)\left\lbrack2x^2+2y^2+2z^2-2xy-2yz-2xz\right\rbrack=0\)

=>\(\left(x+y+z\right)\left\lbrack\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right\rbrack=0\)

mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2>0\) vì x,y,z đôi một khác nhau

nên x+y+z=0

=>y+z=-x

Sửa đề: \(A=2025+\left(y+z\right)^{2025}+x^{2025}\)

\(=2025+\left(-x\right)_{}^{2025}+x^{2025}\)

\(=2025-x^{2025}+x^{2025}=2025\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{2025+2025\cdot2012-2025\cdot2013}\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{2025\left(1+2012-2013\right)}\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{2025\cdot0}\)

\(A=\frac{2013\cdot14+1998+2010\cdot2012}{0}\)

\(A=0\)

=))

\(Y\times6+11\times\dfrac{5}{11}=2025\\ Y\times6+5=2025\\ Y\times6=2025-5\\ Y\times6=2020\\ Y=\dfrac{2020}{6}\\ Y=\dfrac{1010}{3}\)

\(y\times6+11\times\dfrac{5}{11}=2025\\ y\times6+5=2025\\y\times6=2025-5\\ y\times6=2020\\ y=2020:6\\ y=\dfrac{1010}{3}\)

a) (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 2025

(x + x + x + x + x) + (1 + 2 + 3 + 4 + 5)            = 2025

5x + 15                                                             = 2025

5x                                                                     = 2025 - 15

5x                                                                     = 2010

x                                                                       = 2010 : 5

x                                                                       = 402

b) 5 * x - x = 2020

5 * x - x * 1 = 2020

x * (5 - 1)    = 2020

x * 4            = 2020

x                 = 2020 : 4

x                 = 505

mong bạn tick

a) ( x + 1 ) + ( x + 2) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 2025

\(\left(x+x+x+x+x\right)+\left(1+2+3+4+5\right)=2025\)

\(5x+15=2025\)

\(5x=2025-15\)

\(5x=2010\)

  \(x=2010:5\)

  \(x=402\).

b) \(2025^x=9^4\cdot5^4\)

\(\left(45^2\right)^x=\left(9\cdot5\right)^4\)

\(45^{2x}=45^4\)

\(\Rightarrow2x=4\)

\(x=4:2\)

\(x=2\)

Vậy x = 2

=))

\(2025-y\times5=135\\ \Rightarrow y\times5=2025-135\\ \Rightarrow y\times5=1890\\ \Rightarrow y=1890:5=378\)

Bỏ dấu ⇒ đi em, lớp 4 không dùng dấu đó

x+x*4 = 2025
x*(4+1) = 2025
x*5 = 2025 
=> x= 2025:5 = 405

ĐS x=405

x + x . 4 = 2025

x . 1 + x . 4 = 2025

x.(1 + 4) = 2025

       x . 5 = 2025 

            x =  2025 : 5

            x = 405

bạn nhóm ba số giữa vs nhau r lấy x^4+1 xong phân k ra hehe mk cx ko chắc

Đáp án D