\(\frac{x^2-4}{1}+\frac{1}{x-2}=\frac{1}{2}\)

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

\(\Leftrightarrow\frac{x^3-4x-2x^2+8+1}{x-2}=\frac{1}{2}\)

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

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

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

\(\Leftrightarrow x\left(x-2\right)=\frac{x-2}{2\left(x-2\right)}-\frac{18}{2\left(x-2\right)}\)

\(\Leftrightarrow x\left(x-2\right)=\frac{x-20}{2\left(x-2\right)}\)

\(\Leftrightarrow x=\frac{x-20}{2}\)

\(\Leftrightarrow2x=x-20\)

\(\Leftrightarrow2x-x=-20\)

\(\Leftrightarrow x=-20\)

Vậy x=-20

Giúp mình với mình đang cần gấp .-.

\(2x^3+5x^2-3x=0\)

\(x\left(2x^2+5x-3\right)=0\)

\(x\left(2x^2+6x-x-3\right)=0\)

\(x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(x\left(x+3\right)\left(2x-1\right)=0\)

tự làm nốt

hok tốt 

Ta có:

\(2x^3+5x^2-3x=0\)

\(\Rightarrow x\left(2x^2+5x-3\right)=0\)

\(\Rightarrow x\left(2x^2+6x-x-3\right)=0\)

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

\(\Rightarrow x\left(x+3\right)\left(2x-1\right)=0\)

=>x=0 hoặc x=3 hoặc 2x-1=0

=>x=0 hoặc x=3 hoặc x=1/2

Vậy ...

a) \(3x^2+12x-66=0\)

Ta có \(\Delta=12^2+4.3.66=936,\sqrt{\Delta}=6\sqrt{26}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-12+6\sqrt{26}}{6}=-2+\sqrt{26}\\x=\frac{-12-6\sqrt{26}}{6}=-2-\sqrt{26}\end{cases}}\)

b) \(9x^2-30x+225=0\)

Ta có \(\Delta=33^2-4.9.225=-7011\)

\(\Delta< 0\)nên pt vô nghiệm

c) \(x^2+3x-10=0\)

Ta có \(\Delta=3^2+4.10=49,\sqrt{\Delta}=7\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-3+7}{2}=2\\x=\frac{-3-7}{2}=-5\end{cases}}\)

d) \(3x^2-7x+1=0\)

Ta có \(\Delta=7^2-4.3.1=37,\sqrt{\Delta}=\sqrt{37}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+\sqrt{37}}{6}\\x=\frac{7-\sqrt{37}}{6}\end{cases}}\)

a)Vì AB>AC(5cm-1 > 3cm)

=> C nằm giữa A và B

b)K/c từ C đến B là :

(5-1)-3=1(cm)

=> K/c từ C đến D là :

1+6=7(cm)

Vậy k/c từ C đến D là 7 cm.

Nhớ k Cho Mk nha!!!

\(ĐKXĐ:x\inℝ\)

\(\frac{2x}{x^2-x+1}-\frac{x}{x^2+x+1}=\frac{5}{3}\)

\(\Leftrightarrow\frac{2x}{x^2-x+1}-\frac{x}{x^2+x+1}-\frac{5}{3}=0\)

\(\Leftrightarrow\frac{6x\left(x^2+x+1\right)-3x\left(x^2-x+1\right)-5\left(x^4+x^2+1\right)}{3\left(x^4+x^2+1\right)}=0\)

\(\Leftrightarrow6x^3+6x^2+6x-3x^3+3x^2-3x-5x^4-5x^2-5=0\)

\(\Leftrightarrow-5x^4+3x^3+4x^2+3x-5=0\)

\(\Leftrightarrow-5x^4+5^3-2x^3+2x^2+2x^2-2x+5x-5=0\)

\(\Leftrightarrow-5x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-5x^3-2x^2+2x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-5x^3+5x^2-7x^2+7x-5x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[-5x^2\left(x-1\right)-7x\left(x-1\right)-5\left(x-1\right)\right]=0\)

\(\Leftrightarrow-\left(x-1\right)^2\left(5x^2+7x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x^2+7x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x^2+\left(2x+\frac{7}{4}\right)^2+\frac{31}{16}=0\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{1\right\}\)