2x+4x+6x+...+2014x=2015.2016

=>(2+4+6+...+2014).x=2015.2016

tổng trong ngoặc có:(2014-2):2+1=1007(số hạng)

=>tổng= (2014+2).1007:2=1015056

=>1015056.x=4062240

=>x=4030/1007

2x+4x+6x+...+2014x=2015*2016

2x+4x+6x+...+2014x=4062240

x.(2+4+6+...+2014)=4062240

x.1015056=4062240

x=4062240:1015056

x=4,001986097

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

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

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

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

\(\Rightarrow\) \(-3x=0\) 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..

a. Tớ chưa giải ra được, chờ xíu tớ suy nghĩ :>

b. Ta có \(15x^2-11x-14=\left(5x-7\right)\left(3x+2\right)\)

mà \(15x^2-11x-14⋮7\Rightarrow3x+2⋮7\) ( đpcm )

a. Ta có \(x+2014y+y+2014x=2015\left(x+y\right)⋮2015\)

mà \(\left(x+2014y\right)⋮2015\) \(\Rightarrow\left(y+2014x\right)⋮2015\) ( đpcm )

i don't now

mong thông cảm !

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