\(2016^2-16^2=\left(2016+16\right).\left(2016-16\right)=2032.2000=4064000\)

nếu viết thêm chữ số 0 vào bên phải số bé thì được số lớn suy ra số  bé bằng \(\frac{1}{10}\)số lớn 

ta có sơ đồ :

số bé : I-----I

số lớn : I-----I-----I-----I-----I-----I-----I-----I-----I-----I-----I

số bé là :

15576 : [ 1 + 10 ] x 1 = 1416

số lớn là :

1416 x 10 = 14160

b)

\(\left(16x^2-9\right).\left(x+1\right)^2=0\)

\(\Rightarrow\left[\left(4x\right)^2-3^2\right].\left(x+1\right)^2=0\)

=> \(\left(4x\right)^2-3^2=0\) hoặc \(\left(x+1\right)^2=0\)

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

2) \(\left(4x\right)^2-3^2=0\Rightarrow\left(4x+3\right).\left(4x-3\right)=0\)

=> 4x+3 =0 hoặc 4x-3=0

* 4x+3=0 => 4x=-3 => x= \(\dfrac{-3}{4}\)

* 4x-3=0 => 4x= 3 => x= \(\dfrac{3}{4}\)

Vậy x=-1 ; x=0 ; x=\(\dfrac{-3}{4}\) ; x=\(\dfrac{3}{4}\)

like nha ae !!!!!!!!!!!!!!!!!!

a)

\(\Rightarrow\left[\left(x^3+3^3\right)+\left(x+3\right)\right].\left(x-9\right)=0\)

\(\Rightarrow\left[\left(x+3\right)\left(x^2-3x+3^2\right)+\left(x+3\right)\right].\left(x-9\right)=0\)

\(\Rightarrow\left[\left(x+3\right).\left(x^2-3x+9+1\right)\right].\left(x-9\right)=0\)

\(\Rightarrow\left[\left(x+3\right).\left(x^2-3x+10\right)\right].\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)=0\) hoặc x-9=0 hoặc \(x^2-3x+10=0\)

1) x-9=0 => x=9

2) x+3 = 0 => x= -3

3) \(x^2-3x+10=0\Rightarrow x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}=0\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=0\Rightarrow\left(x-\dfrac{3}{2}\right)^2=-\dfrac{31}{4}\left(ktm\right)\)

Vậy x=9 ; x=-3

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

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

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)

\(\Rightarrow\left(x+2016\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)

do \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)

\(\Rightarrow x+2016=0\Rightarrow x=2016\)

váy x=2016

câu 1

f(x) = \(x^2+x+1\)

g(x)=\(x^4+x^3+x^2+x+1\)

\(\Rightarrow f\left(x\right).g\left(x\right)=\left(x^2+x+1\right)\left(x^4+x^3+x^2+x+1\right)\)

\(=x^6+x^5+x^4+x^3+x^2+x^5+x^4+x^3+x^2+x+x^4+x^3+x^2+x+1\)

\(=x^6+2x^5+3x^4+3x^3+3x^2+x+1\)

=> f(x).g(x) có bậc là 6

Câu 2

ta có \(A=x^2-2x-1\Rightarrow A=x^2-2x+1-2\)

\(A=\left(x-1\right)^2-2\ge-2\)

dấu bằng xảy ra khi A đạt GTNN bằng -2

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

Vây A đạt GTNN là -2 tại x=1

like nha ae !!!!

ta có \(35^{2005}-35^{2004}=35^{2004}.35-35^{2004}=35^{2004}.\left(35-1\right)=35^{2004}.34\)

do \(34⋮17\Rightarrow35^{2004}.34⋮17\left(đpcm\right)\)