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, $\frac{x-2}{x-m}=\frac{x-1}{x+2}$ 

⇒ $( x-2).( x+2)= ( x-1).( x-m)$

⇔ $x²-4= x²-mx-x+m$

⇔ $x+mx-m-4= 0$

⇔ $x.( m+1)= m+4$

⇔ $x= \frac{m+4}{m+1}$ 

Đáp án:

 

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