\(x^{15}=x\)

\(\Rightarrow x^{15}-x=0\)

\(\Rightarrow x.\left(x^{14}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^{14}=0+1=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Ta có : x : x¹⁵ = x 

<=> x = x¹⁶

=> x = 1 ; 0 

X=4/15:2/5=2/3

2/3

X = 4/5 : 2/15

X = 2/3

a: \(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

hay \(x\in\left\{0;\sqrt{3};-\sqrt{3}\right\}\)

b: \(=\dfrac{x^3-3x^2+6x-8}{x-2}=\dfrac{x^2-2x-x^2+2x+4x-8}{x-2}=x^2-x+4\)

a, = 76 + 30 - 5

= 101

b, x - 2,751 = 15,12

x = 15,12 + 2,751 

x = 17,871

\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)

\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)

\(\Leftrightarrow-57x=-171\)

\(\Leftrightarrow x=3\)

\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)

\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )

\(\Leftrightarrow x=-2016\)

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

x:x*15-3,7=2,3

x:2*15-3,7=2,3

x:2*15=2,3+3,7

x:2*15=6

x:2=6:15

x*2=0,4

x=0,4:2

x=0,2

`x+x:0,1+x:0,01=111`

`x+x.10+x.100=111`

`x.(1+10+100)=111`

`111x=111`

`x=1`

Vậy x cần tìm là `1`

x + x : 0,1 + x : 0,01 = 111

x.1+x.10+x.100=111

x.(1+10+100)=111

x.111=111

x=111:111

x=1

#YM

x + x : 0,1 + x : 0,01 = 111

x.1+x.10+x.100=111

x.(1+10+100)=111

x.111=111

x=111:111

x=1

                                                      THAM KHẢO !