x+2.x+3.x+4.x+....+2012.x=2012.2013

x(1+2+3+4+...+2012)=2012.2013

x(2012.2013:2)=2012.2013

x=2012.2013:(2012.2013:2)

x=2

x+2.x+3.x+4.x+....+2012.x=2012.2013

x.(1+2+3+4+...+2012)=2012.2013

      Số số hạng từ 1 đến 2012 là:

          (2012-1):1+1=2012(số hạng)

       Tổng từ 1 đến 2012 là:

           (2012+1)x2012:2=2025078

x.(1+2+3+4+...+2012)=2012.2013

x.2025078=4050156

x=4050156:2025078

x=2

Vậy x=2

<=> x*(1+2+3+4+...+2012) = 2012*2013

<=> x*1/2 *2012*2013 = 2012*2013

<=> x*1/2 = 1

<=> x = 2.

\(Bài.1:\\ a,104^2-16=104^2-4^2=\left(104+4\right)\left(104-4\right)=108.100=10800\\ b,9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\\ =\left(9.2\right)^8-\left(18^8-1\right)=18^8-18^8+1=1\\ c,999^3+3.999^2+3.999+1\\ =999^3+3.999^2.1+3.999.1^2+1^3=\left(999+1\right)^3=1000^3=1000000000\\ d,42^3-6.42^2+12.42-8\\ =42^3-3.42^2.2+3.42.2^2-2^3\\ =\left(42-2\right)^3=40^3=64000\)

Bài 1

a) 104² - 16

= 104² - 4²

= (104 - 4)(104 + 4)

= 100.108

= 10800

b) 9⁸.2⁸ - (18⁴ - 1)(18⁴ + 1)

= 18⁸ - (18⁸ - 1)

= 18⁸ - 18⁸ + 1

= 1

c) 999³ + 3.999² + 3.999 + 1

= (999 + 1)³

= 1000³

= 1000000000

d) 42³ - 6.42² + 12.42 - 8

= (42 - 2)³

= 40³

= 64000

Bài 2

a) x(x - 2012) - 2013x + 2012.2013 = 0

⇔ x(x - 2012) - 2013(x - 2012) = 0

⇔ (x - 2012)(x - 2013) = 0

⇔ x - 2012 = 0 hoặc x - 2013 = 0

*) x - 2012 = 0

⇔ x = 2012

*) x - 2013 = 0

⇔ x = 2013

Vậy x = 2012; x = 2013

b) (x - 1)³ + 1 + 3x(x - 4) = 0

⇔ x³ - 3x² + 3x - 1 + 1 + 3x² - 12x = 0

⇔ x³ - 9x = 0

⇔ x(x² - 9) = 0

⇔ x(x - 3)(x + 3) = 0

⇔ x = 0 hoặc x - 3 = 0 hoặc x + 3 = 0

*) x - 3 = 0

⇔ x = 3

*) x + 3 = 0

⇔ x = -3

Vậy x = -3; x = 0; x = 3

c) (x + 4)² - 16 = 0

⇔ (x + 4)² - 4² = 0

⇔ (x + 4 - 4)(x + 4 + 4) = 0x

⇔ (x + 8) = 0

⇔ x = 0 hoặc x + 8 = 0

*) x + 8 = 0

⇔ x = -8

Vậy x = -8; x = 0

nốt ý b:

\(\left(x-1\right)^3+1+3x\left(x-4\right)=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+1+3x^2-12x=0\)

\(\Leftrightarrow x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Vậy ..............

\(a,x\left(x-2012\right)-2013x+2012.2013=0\)

\(=x\left(x-2012\right)+2013\left(-x+2012\right)=0\)

\(\Rightarrow x\left(x-2012\right)-2013\left(x-2012\right)=0\)

\(\Rightarrow\left(x-2013\right)\left(x-2012\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2013=0\\x-2012=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2013\\x=2012\end{matrix}\right.\)

Vậy...

=>20x+190=950

=>20x=760

hay x=38

`20x+190=950`

`20x=760`

`x= 760: 20`

`x= 38`

20 x X = 950 - (1 + 2 + 3 + 4 + 5 + ... + 19)

20 x X = 950 - 190

20 x X = 760

X = 760 : 20

X = 38

 x³ - x² - x = 1/3 
<=> x³ = x² + x + 1/3 
<=> 3x³ = 3(x² + x + 1/3) 
<=> 3x³ = 3x² + 3x + 1 
<=> 3x³ + x³ = x³ + 3x² + 3x + 1 
<=> 4x³ = (x + 1)³ 
<=> ³√(4x³) = ³√(x + 1)³ 
<=> ³√4.x = x + 1 
<=> ³√4.x - x = 1 
<=> x(³√4 - 1) = 1 
<=> x = 1/(³√4 - 1) 

Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

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

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow x^2-x^2+2x=-4+3\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\frac{1}{2}\)

Vậy \(x=-\frac{1}{2}\)

\(x^2\left(x+1\right)+\left(x+1\right)=y^3\)

\(\left(x+1\right)\left(x^2+1\right)=y^3\)

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

\(\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x^2=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\kothoaman\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=-1\\y^3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=0\end{cases}}\)

Vậy x = -1, y =0

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x+\frac{1}{6}=0\)  

\(\Rightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)

k cho minh

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)

\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x-\frac{1}{6}=0\)

\(\Leftrightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=0\)

Tính ra nhé !

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=x+\frac{1}{6}\)

\(\Rightarrow4x+\frac{77}{60}=x+\frac{1}{6}\)

\(\Rightarrow3x=\frac{1}{6}-\frac{77}{60}\)

\(\Rightarrow3x=-\frac{67}{60}\)

\(\Rightarrow x=-\frac{67}{60}\div3=\frac{-67}{60.3}=-\frac{67}{180}\)

Vậy x = .........

a) x+2x+3x+4x+...+2011x = 2012.2013

\(\Rightarrow\) x(1+2+3+4+...+2011) = 4050156

\(\Rightarrow\) x.2023066 = 4050156

\(\Rightarrow\) x = 4026/2011

Câu a ko nhất thiết phải tính ra số lớn như thế đâu