1) \(x^2+6x+9\)

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

2) \(10x-25-x^2\)

\(=-25+10x-x^2\)

\(=-\left(5-x\right)^2\)

3) \(8x^3-\dfrac{1}{8}\)

\(=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\)

\(=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

4) \(\dfrac{1}{25}x^2-64y^2\)

\(=\left(\dfrac{1}{5}x\right)^2-\left(8y\right)^2\)

\(=\left(\dfrac{1}{5}x+8y\right)\left(\dfrac{1}{5}x-8y\right)\)

\(x^2+6x+9=\left(x+3\right)^2\)

\(10x-25-x^2=-\left(x-5\right)^2\)

\(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

mình học lớp 5 mà

\(f\left(x\right)=x^6-10x^5+10x^4-10x^3+10x^2-10x+10\)

\(f\left(x\right)=x^5\left(x-10\right)+x^3\left(x-10\right)+x\left(x-10\right)+10\)

\(f\left(x\right)=\left(x-10\right)\left(x^5+x^3+x\right)+10\)

\(f\left(x\right)=x\left(x-10\right)\left(x^4+x^2+1\right)+10\)

\(\Rightarrow f\left(9\right)=9.\left(9-10\right)\left(9^4+9^2+1\right)+10\)

\(\Leftrightarrow f\left(9\right)=9.\left(-1\right).\left(6643\right)+10\)

\(\Leftrightarrow f\left(9\right)=-59777\)

P/s : làm cho zui thôi nha , sai đừng đáp đá 

\(x=9\)\(\Rightarrow x+1=10\)

\(\Rightarrow f\left(9\right)=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)

               \(=x^6-x^6-x^5+x^5+.......-x+x+1=1\)

\(f\left(x\right)=10x+5\)

Nghiệm của \(f\left(x\right)\) là \(x\Leftrightarrow10x+5=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) (1)

\(g\left(x\right)=x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}\)

Nghiệm của \(g\left(x\right)\)là \(x\Leftrightarrow x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}=0\)

\(\Leftrightarrow x^2\left(x+\dfrac{1}{2}\right)+3\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x^2+3=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) (2)

Từ (1) và (2) => ĐPCM

\(x\) + \(\dfrac{9}{10}\) = \(\dfrac{16}{10}\)

\(x\)           = \(\dfrac{16}{10}\) - \(\dfrac{9}{10}\)

\(x\)           = \(\dfrac{7}{10}\)

\(x\)    =0,7

Lời giải:

Vì $x=9$ nên $x-9=0$
Ta có:

$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$

$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$

$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$

$=x-10=9-10=-1$

e) \(E=x^5-15x^4+16x^3-29x^2+13x\) tại x = 14

\(E=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+x\left(x-1\right)\)

\(E=x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)

\(E=-x\)

\(E=-14\)

d)  \(D=x^3-30x^2-31+1\) tại x = 31

\(D=31^3-30.31^2-31+1\)

\(D=31^2\left(31-30-1\right)+1\)

\(D=0+1\)

\(D=1\)

a) f(10x)=10f(x) 

Ta có:

y=f(x)=kx

\(\Rightarrow\)f(10x)=k10x=10kx (*)

\(\Rightarrow\)10f(x)=10kx (**)

Từ (*) và (**)

\(\Rightarrow\)f(10x)=10f(x) 

\(\Rightarrow\)đpcm

b) f(x₁+x₂)=f(x₁)+f(x₂)

Ta có:

y=f(x) =kx

\(\Rightarrow\)f(x₁+x₂)=k(x₁+x₂) (*)

\(\Rightarrow\)f(x₁)+f(x₂)=kx₁+kx₂=k(x₁+x₂) (**)

Từ (*) và (**)

\(\Rightarrow\)f(x₁+x₂)=f(x₁)+f(x₂)

\(\Rightarrow\)đpcm

c) f(x₁-x₂)=f(x₁)-f(x₂)

Ta có:

y=f(x)=kx

\(\Rightarrow\)f(x₁-x₂)=k(x₁-x₂) (*)

\(\Rightarrow\)f(x₁)-f(x₂)=kx₁-kx₂=k(x₁-x₂) (**)

Từ (*) và (**)

\(\Rightarrow\)f(x₁-x₂₎=f(x₁-x₂)

\(\Rightarrow\)đpcm

P/s: đã sửa đề