Câu 1: 

(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0

=> 2x + 1 + 2x + 2 + ... + 2x + 2015 = 0

=> 2015.2x + (1 + 2 + ... + 2015) = 0

=> 4030x + (2015 + 1).2015:2 = 0

=> 4030x + 2031120 = 0

=> x = -504

Câu 2:

x - y = 8; y - z = 10; x + z = 12

=> (x - y) + (y - z) = 8 + 10 = 18

=> x - z = 18

=> x = (12 + 18) : 2 = 15

=> z = 15 - 18 = -3

=> y = 15 - 8 = 7

=> x + y + z = 15 + 7 + (-3) = 19

a, -504

b,19 dung thi tic minh nha

Đặt A = ( 2x + 1 ) + ( 2x + 2 ) + .... + ( 2x + 2015 ) = 0

=> A = ( 2x + 2x + 2x + ..... + 2x ) + ( 1 + 2 + 3 + ..... + 2015 )

                   Tổng 1                                 tổng 2

Số các số hạng của 2 tổng là : ( 2015 - 1 ) : 1 + 1 = 2015 ( số hạng )

Tổng 1 : 2x . 2015 = 4030x

Tổng 2 : ( 2015 + 1 ) . 2015 : 2 = 2031120

=> 4030x + 2031120 = 0 => 4030x = - 2031120 => x = - 2031102 : 4030 => x = - 504

Vậy x = - 504

(\(x-1,25\))(\(x-8\)) = 0

\(\left[{}\begin{matrix}x-1,25=0\\x-8=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1,25\\x=8\end{matrix}\right.\)

Vậy \(x\) \(\in\) { 1,25; 8}

Ta có : x - 1,25 = 0      hoặc        x - 8 = 0

           x            = 0 + 1,25           x      =  0 + 8

           x            =  1,25                x      =  8

  Vậy : x = 1,25 hoặc x = 8

($x-1,25$)($x-8$) = 0

[�−1,25=0�−8=0[x−1,25=0x−8=0​

[�=1,25�=8[x=1,25x=8​

Vậy $x$ $\in$ { 1,25; 8}

3 commercial

4 exhibition

5 humorous

6 arrangement

7 demonstrate

8 emigration

9 successfully

10 unsuccessful

11 assistant

12 speech

13 agreement

14 orphan

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

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

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

\(=\dfrac{3}{x}\cdot\left(-x\right)-\dfrac{3}{x}\cdot\dfrac{3}{3-x}=-3+\dfrac{9}{x\left(x-3\right)}\)

\(=\dfrac{-3x\left(x-3\right)+9}{x\left(x-3\right)}=\dfrac{-3x^2+9x+9}{x\left(x-3\right)}\)

cau này mình làm rồi nhưng ko nhớ

   :))))))))))))))))

à

cái này dễ

\(2x^2+6x-8=0\)

<=> \(2x^2-2x+8x-8=0\)

<=> \(2x\left(x-1\right)+8\left(x-1\right)=0\)

<=> \(\left(2x+8\right)\left(x-1\right)=0\)

<=> \(\hept{\begin{cases}2x+8=0\\x-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-4\\x=1\end{cases}}\)

\(2x^2-x-1=0\)

<=> \(2x^2-2x+x-1=0\)

<=> \(2x\left(x-1\right)+\left(x-1\right)=0\)

<=> \(\left(2x+1\right)\left(x-1\right)=0\)

<=> \(\hept{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

\(4x^2-5x-9=0\)

<=> \(4x^2+4x-9x-9=0\)

<=> \(4x\left(x+1\right)-9\left(x+1\right)=0\)

<=> \(\left(4x-9\right)\left(x+1\right)=0\)

<=> \(\hept{\begin{cases}4x-9=0\\x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)

học tốt

\(2x^2+6x-8=0\)

\(< =>2x^2-2x+8x-8=0\)

\(\Leftrightarrow2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)

\(\Leftrightarrow2x+8=0\)hoặc \(x-1=0\)

\(\Leftrightarrow x=-4\)hoặc \(x=1\)

câu t2 bn tách -x thành -2x+x nha rồi làm tương tự câu trên á

câu t3 tách -5x thành +4x-9x nha rồi làm tương tự câu trên á

Chúc bn học tốt !!

^.^

thằng Lê Quang Trung lắm mồm