hom nay 1h  mih thi huyen giup voi nha

Nếu ở cuối là = 2013.1007 thì phương trình có nghiệm là x = 1 Vì:

Ta thực hiện thao tác nhóm như sau:

(x+2012x) + (2x+2011x) + (3x+2010x) + ....+ 2013x = 2013.1007

 2013x + 2013x + 2013x + ...+ 2013x = 2013.1007

có tất cả 1006 số 2013x cộng với 2013x = 2013.1007

Vậy: 1006.2013x  + 2013x = 2013.1007

 (1006 + 1)2013x = 2013.1007  x = (2013.1007)/[(1006+1)2013] = 1

X=1

Ko sai

Bai nay lam  roi

Ap dung ct tinh tong so hang cua 1 day so

roi the vao

\(A.x^2-16x=0\)

\(x^2-\left(4x\right)^2=0\)

\(\left(x-4x\right)\left(x+4x\right)=0\)

\(\left(-3x\right)\left(5x\right)=0\)

\(\Rightarrow\) \(-3x=0\) hoặc \(5x=0\)

\(x=\dfrac{0}{-3}\) hoặc \(x=\dfrac{0}{5}\)

Vậy \(x=0\) hoặc \(x=0.\)

B. 4x² - 4x + 1 = 0

(2x)² - (2x)² + 1² = 0

(2x - 2x + 1 ) (2x + 2x +1) = 0

1 (4x + 1) =0

=> 1 (4x + 1) =0

4x + 1 = 0

4x = 0-1

Vậy x = \(\dfrac{-1}{4}.\)

C. (3x-1)² - (2x+3)² = 0

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

(x + 2)(5x + 2) = 0

=> x + 2 =0 hoặc 5x + 2 =0

x = 0 - 2 hoặc 5x = 0 - 2

Vậy x = -2 hoặc x = \(\dfrac{-2}{5}.\)

Còn về câu d thì mình hơi phân vân, tại mình dốt toán lắm

a/ \(x^2-16x=0\)

\(\Leftrightarrow x\left(x-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)

Vậy...

b/ \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy..

c/ \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1-2x-3\right)\left(3x-1+2x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy...

d/ \(2013x^2-2014x+1=0\)

\(\Leftrightarrow2013x^2-x-2013x+1=0\)

\(\Leftrightarrow x\left(2013x-1\right)-\left(2013x-1\right)=0\)

\(\Leftrightarrow\left(2013x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2013x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2013}\\x=1\end{matrix}\right.\)

Vậy..

bằng 1 bạn nhé

x+2x+3x+.....+2013x=2013.2017

=>x.(1+2+3+.....+2013)=2013.2017

=>\(x.\frac{2013.\left(2013+1\right)}{2}=2013.2017\Rightarrow x.2027091=4060221\Rightarrow x=\frac{2017}{1007}\)

x+2x+3x+...+2013x=2013.2017

=>x(1+2+3+...+2013)=2013.2017
 

=>X\(\frac{2013\left(2013+1\right)}{2}=2027091=4060221\)

=>X.2027091=4060221

X=4060221:2027091=2017/1007

=>x.(1+2+3+..+2013)=2013

=>x. [2013.(2013+1)]/2=2013

=>x.2027091=2013

=>x=2013/2027091=1/1007