\(x^2-3x=0\)

\(\Rightarrow x\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy..

trả lời

x^2+3x=0

<=>x(x+3)=0

th1:x=0

th2:x+3=0=>x=-3

vậy x=0 hoặc -3

ĐKXĐ: \(x\ne\pm2\)

\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=\frac{-7}{23}\left(tm\right)\)

Vậy: \(S=\left\{-\frac{7}{23}\right\}\)

=.= hk tốt!!

Giải :

\(\text{ĐKXĐ}\: :\: x\ne\pm2\)

\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

 \(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

Khử mẫu : \(\left(-6x^2-12x+x+2\right)+\left(9x^2-18x+4x-8\right)=3x^2-2x+1\)

           \(\Leftrightarrow-23x=7\Leftrightarrow x=\frac{7}{23}\).

TL:

\(x^2-6x+9-4>0\) 

\(\left(x-3\right)^2-4>0\)

\(\left(x-3\right)^2-2^2>0\) 

\(\left(x-3+2\right)\left(x-3-2\right)>0\)

(x-1)(x-5)>0

=>x>5

vậy.......

hc tốt

thêm 1 trường hợp:

x<1 nha

chúc bn

hc tốt

\(x^2-6x+5>0\)

\(\Leftrightarrow x^2-x-5x+5>0\)

\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}x-1>0\\x-5>0\end{cases}}\) hoặc   \(\hept{\begin{cases}x-1< 0\\x-5< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>1\\x>5\end{cases}}\)   hoặc   \(\hept{\begin{cases}x< 1\\x< 5\end{cases}}\)

Vậy : x > 5 hoặc x < 1

=.= hk tốt!!

Nghiệm của bất phương trình đc biểu diễn trên trục số 

0 5 10 15 20 25 30 ( xấu quá thong cảm )

Kết quả thu đc :x\(\in\){1}U{5}

ko chắc lắm 

hc tốt

\(a\left(b^2+c^2+bc\right)+b\left(c^2+a^2+ac\right)+c\left(a^2+b^2+ab\right)\)

\(=ab^2+ac^2+abc+bc^2+ba^2+abc+ac^2+bc^2+abc\)

\(=c^2\left(b+a\right)+\left(b^2+3\text{a}b+a^2\right)c+ab^2+a^2b\)

\(=bc^2+ac^2+b^2c+3\text{a}bc+a^2c+ab^2+a^2b\)

\(=\left(c+b+a\right)\left(bc+ac+ab\right)\)

Với n lẻ thì: \(^{a^n}\)+ \(^{b^n}\) = ( a+ b)*(\(^{a^{n-1}}\)- \(^{a^{n-2}}\) * \(^{b+a^{n-3}}\) * \(^{b^2}\)-........-\(^{a\cdot b^{n2}}\)+ \(^{b^{n-1}}\))

hay:\(^{a^n}\)+ \(^{b^n}\) chia hết cho  a+b

\(^{1^n}\)+ \(^{2^n}\)+\(^{3^n}\) + \(^{4^n}\)= ( \(^{1^n}\)+ \(^{4^n}\)) +(\(^{2^n}\)+ \(^{3^n}\))

 Vậy với n lẻ \(^{1^n}\)+ \(^{4^n}\) và  \(^{2^n}\) + \(^{3^n}\) đều chia hết cho 5 nên N lẻ

a) \(3n+5⋮n+4\)

\(\Rightarrow3.\left(n+4\right)-7⋮n+4\)

Mà \(3.\left(n+4\right)⋮n+4\)

\(\Rightarrow7⋮n+4\)

Tự tìm nốt

b) \(n^2+5⋮n+1\)

\(\Rightarrow n^2+n-n+5⋮n+1\)

\(\Rightarrow n.\left(n+1\right)-\left(n-5\right)⋮n+1\)

mà \(n.\left(n+1\right)⋮n+1\)

\(\Rightarrow n-5⋮n+1\)

\(\Rightarrow n+1-6⋮n+1\)

mà \(n+1⋮n+1\)

\(\Rightarrow6⋮n+1\)

Tìm nốt