Question

[Sadness30]

Giải các Phương trình sau ;

a) (x² - 2x +1) - 4 = 0

b) I5x-5l  = 0

c)  \(\frac{x-2}{x+2}\) + \(\frac{3}{x-2}\) = \(\frac{x^2-11}{x^2-4}\)

Answer 1

[๖²⁴ʱ๖ۣۜTɦủү❄吻༉]

a, \(\left(x^2-2x+1\right)-4=0\)

\(x^2-2x+1-4=0\)

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

\(\Delta=b^2-4ac=\left(-2\right)^2-4.1.3=4-12=-8< 0\)

Nên pt vô nghiệm 

b, \(\left| 5x-5\right|=0\)

\(\Leftrightarrow5x-5=0\Leftrightarrow5x=5\Leftrightarrow x=1\)

c, ĐKXĐ : \(\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2-4\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\x\ne2\\x\ne\pm2\end{cases}\Rightarrow}x\ne\pm2}\)

\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)

\(\frac{\left(x-2\right)^2\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)\left(x^2-4\right)}+\frac{3\left(x+2\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)\left(x^2-4\right)}=\frac{\left(x^2-11\right)\left(x+2\right)\left(x-2\right)}{\left(x^2-4\right)\left(x+2\right)\left(x-2\right)}\)

\(\left(x-2\right)^2\left(x^2-4\right)+3\left(x+2\right)\left(x^2-4\right)=\left(x^2-11\right)\left(x+2\right)\left(x-2\right)\)

\(\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)

\(x^2-x+10=x^2-11\)

\(x^2-x+10-x^2+11=0\)

\(-x+21=0\Leftrightarrow x-21=0\Leftrightarrow x=21\)Theo ĐKXĐ : => tm 

Answer 2

[Kiyotaka Ayanokoji]

a, \(\left(x^2-2x+1\right)-4=0\) \(\Leftrightarrow\left(x-1\right)^2=4=\left(\pm2\right)^2\)

                                                           \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy phương trình có 2 nghiệm x=(3; -1)

b, \(\left|5x-5\right|=0\Leftrightarrow5x-5=0\)

                                 \(\Leftrightarrow5x=5\Rightarrow x=1\)

Vậy phương trình có nghiệm x=1

c, \(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)\(\left(x\ge0;x\ne2\right)\) \(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right).\left(x+2\right)}+\frac{3.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right).\left(x+2\right)}\)

                                                                  \(\Leftrightarrow\left(x-2\right)^2+3.\left(x+2\right)=x^2-11\)

                                                                 \(\Leftrightarrow x^2-4x+4+3x+6=x^2-11\)

                                                                 \(\Leftrightarrow x=21\left(TM\right)\)

Vậy phương trình có nghiệm x=21