Bài 4: Điều kiện: $x≠a$
a, $\frac{x-1}{x+a}-\frac{x}{x-a}= \frac{x+2a}{a²-x²}$
⇔ $\frac{( x-1).(a-x)}{(a+x).(a-x)}+\frac{x.( a+x)}{( a-x).( a+x)}= \frac{x+2a}{(a-x).(a+x)}$
⇔ $\frac{( x-1).(a-x)+x.(a+x)}{(a+x).(a-x)}= \frac{x+2a}{(a-x).(a+x)}$
⇔ $( x-1).(a-x)+x.(a+x)= x+2a$
⇔ $ax-x²-a+x+ax+x²= x+2a$
⇔ $2ax= 3a$
Nếu a= 2 thì $2.2.x= 3.2$
⇔ $x=\frac{3}{2}$
b, Phương trình có nghiệm x= 1
⇒ $2a= 3a$
⇔ $a= 0$
Bài 5:
a, $x²-x-18+\frac{72}{x²-x}= 0$
Đặt $x²-x= t$
⇒ $t-18+\frac{72}{t}= 0$
⇔ $t²-18t+72= 0$
⇔ $t²-12t-6t+72= 0$
⇔ $( t-12).( t-6)= 0$
⇔ \(\left[ \begin{array}{l}t= 12\\t=-6\end{array} \right.\)
( Đến đây bạn thay $t= x²-x$ để tính x nhé)
b, $\frac{1}{x.( x+1)}+\frac{1}{( x+1).( x+2)}+\frac{1}{( x+2).(x+3)}= \frac{3}{10}$
⇔ $\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}= \frac{3}{10}$
⇔ $\frac{1}{x}-\frac{1}{x+3}= \frac{3}{10}$
⇔ $\frac{3}{x.( x+3)}= \frac{3}{10}$
⇔ $x.( x+3)= 10$
⇔ $x²+3x-10= 0$
⇔ $( x-2).( x+5)= 0$
⇔ \(\left[ \begin{array}{l}x= 2\\x= -5\end{array} \right.\)
c, $\frac{2}{x²+4x+3}+\frac{5}{x²+11x+24}+\frac{2}{x²+18x+80}= \frac{9}{52}$
⇔ $\frac{2}{( x+1).( x+3)}+\frac{5}{( x+3).( x+8)}+\frac{2}{( x+8).( x+10)}= \frac{9}{52}$
⇔ $\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}= \frac{9}{52}$
⇔ $\frac{1}{x+1}-\frac{1}{x+10}= \frac{9}{52}$
⇔ $\frac{9}{( x+1).( x+10)}= \frac{9}{52}$
⇔ $( x+1).( x+10)= 52$
⇔ $x²+11x-42= 0$
⇔ $( x-3).( x+14)= 0$
⇔ \(\left[ \begin{array}{l}x= 3\\x= -14\end{array} \right.\)
d, $x²+\frac{4x²}{( x+2)²}= 12$
e, $\frac{1}{x-1}+\frac{2x²-5}{x³-1}= \frac{4}{x²+x+1}$
⇔ $\frac{x²+x+1}{x³-1}+\frac{2x²-5}{x³-1}= \frac{4.( x-1)}{x³-1}$
⇒ $x²+x+1+2x²-5= 4.( x-1)$
⇔ $3x²-3x= 0$
⇔ $x.( x-1)= 0$
⇔ \(\left[ \begin{array}{l}x= 0\\x= 1 ( Loại)\end{array} \right.\)
f, $x²+( \frac{x}{x+1})²= \frac{5}{4}$
⇒ $4x².( x+1)²+4x²= 5.( x+1)²$
⇔ $4x².( x²+2x+1)= x²+10x+5$
⇔ $4x^{4}+8x³+3x²-10x-5= 0$
⇔ \(\left[ \begin{array}{l}x= -0,5\\x= 1 ( Loại)\end{array} \right.\)
g, $\frac{x-2}{x-m}=\frac{x-1}{x+2}$
⇒ $( x-2).( x+2)= ( x-1).( x-m)$
⇔ $x²-4= x²-mx-x+m$
⇔ $x+mx-m-4= 0$
⇔ $x.( m+1)= m+4$
⇔ $x= \frac{m+4}{m+1}$
