\(\frac{x^2-x-6}{x-3}=\frac{x^2-3x+2x-6}{x-3}=\frac{x\left(x-3\right)+2\left(x-3\right)}{\left(x-3\right)}=x+2=0\Leftrightarrow x=-2\)

\(\frac{x^2+2x-\left(3x+6\right)}{x+2}=\frac{x\left(x+2\right)-3\left(x+2\right)}{x+2}=x-3=0\Leftrightarrow x=3\)

\(\frac{4}{x-2}-\left(x-2\right)=0\Leftrightarrow\frac{4}{a}-a=0\left(a=x-2\right)\Leftrightarrow\frac{4}{a}=a\Leftrightarrow a^2=4\Leftrightarrow a=\pm2\Leftrightarrow x=4\text{ hoặc 0}\)

a) ĐKXĐ: x \(\ne\)3

Ta có: \(\frac{x^2-x-6}{x-3}=0\)

<=> x² - x - 6 = 0

<=> x² - 3x + 2x - 6 = 0

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

<=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=3\left(vn\right)\end{cases}}\)

Vậy S = {-2}

b) ĐKXĐ: x \(\ne\)-2

Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x+2}=0\)

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

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

<=> \(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-2\left(vn\right)\end{cases}}\)

Vậy S = {3}

c) ĐKXĐ: x \(\ne\)2

Ta có: \(\frac{4}{x-2}-x+2=0\)

<=> \(\frac{4-\left(x-2\right)^2}{x-2}=0\)

<=> \(\left(2-x+2\right)\left(2+x-2\right)=0\)

<=> \(x\left(4-x\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\4-x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy S = {0; 4}

1,2−(x−1,4)=−6(x+0,9)

<=> 1,2 - x + 1,4 = -6x -6.0,9

<=> 2,6 - x = -6x - 5,4

<=> 6x - x  + 2,6 + 5,4 =0 

<=> 5x + 8 = 0 

a) Với a = 5 thì b = 8

b) Nghiệm của phương trình là -8/5

\(\left(x^2-1+x\right)\left(x2-1+3x\right)-x\)

\(\Leftrightarrow\left(x^2-1+x\right)\times\left(5x-1\right)-x\)

\(\Leftrightarrow5x^3-x^2-5x+1+5x^2-x-x\)

\(\Leftrightarrow5x^3+4x^2-7x+1\)

Mình đã rút gọn ngắn nhất có thể rồi đấy!

a)A=\(x^2-6x+2=x^2-2.3x+9-7\)\(=\left(x-3\right)^2-7\ge-7\)với mọi x\(\inℝ\)

Dấu bằng xảy ra\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy minA = - 7 tại x = 3

b)\(B=4x^2-x+2=4x^2-2.2x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+2\)

  \(=\left(2x-\frac{1}{4}\right)^2+\frac{31}{16}\ge\frac{31}{16}\)với mọi x\(\inℝ\)

Dấu bằng xảy ra \(\Leftrightarrow2x-\frac{1}{4}=0\Leftrightarrow x=\frac{1}{8}\)

Vậy minB = \(\frac{31}{16}\)tại \(x=\frac{1}{8}\)

giúp tui vs mọi ng ơi

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

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

\(\Leftrightarrow x^3-3x^2+3x=x^2\)

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

\(\Leftrightarrow x^3-4x^2+3x=0\)

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

\(\Leftrightarrow x\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-3=0\\x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=3\\x=1\end{cases}}}\)

Vậy \(x\in\left\{0;3;1\right\}\)

Bạn tự vé hình nhé!
Xét \(\Delta\)ABD có: OM//AB (gt) => \(\frac{OM}{AB}=\frac{DO}{DB}\left(1\right)\)

Xét \(\Delta\)ABC có: ON //AB (gt) => \(\frac{ON}{AB}=\frac{CO}{CA}\left(2\right)\)

Mặt khác: AB//CD (gt) =>\(\frac{DO}{DB}=\frac{CO}{CA}\left(3\right)\)

(1)(2)(3) => \(\frac{OM}{AB}=\frac{ON}{AB}\)=> OM=ON (đpcm)

Nguồn: loigiaihay.com

\(\left(x^2-2x-1\right)\cdot\left(x-3\right)\)

\(=x^2\cdot x-2x\cdot x-1\cdot x-x^2\cdot3+2x\cdot3+1\cdot3\)

\(=x^3-2x^2-1-3x^2+6x+3\)

\(=x^3-5x^2+6x+2\)

\(\left(x^2-2x-1\right)\left(x-3\right)\)

\(\Leftrightarrow x^3-3x^2-2x^2+6-x+3\)

\(\Leftrightarrow x^3-5x^2-x+9\)

\(\Leftrightarrow x\left(x^2-5x-1\right)+9\)