Đáp án:
Giải thích các bước giải:Bài 18
a) ĐKXĐ x$\neq$ 3/5;x$\neq$ -3/5
ta có $\frac{4}{-25x^2+20x-3}$ =$\frac{3}{5x-1}$ -$\frac{2}{5x-3}$
⇔$\frac{4}{(5x-1)(5x-3)}$ =$\frac{3(5x-3)}{(5x-1)(5x-3)}$ -$\frac{2(5x-1)}{(5x-1)(5x-3)}$
⇒4-3(5x-3)+2(5x-1)=0
⇔4-15x+9+10x-2=0
⇔-5x=11
⇔x=-11/5(TMĐK)
Vậy x=-11/5
b) ĐKXĐ x$\neq$ 1;x$\neq$ 2;x$\neq$ 3
Ta có $\frac{1}{x^2-3x+2}$ +$\frac{1}{x^2-5x+6}$ =$\frac{2}{x^2-4x+3}$
⇔$\frac{1}{(x-2)(x-1)}$ +$\frac{1}{(x-3)(x-2)}$ =$\frac{2}{(x-3)(x-1)}$
⇔$\frac{1(x-3)}{(x-1)(x-2)(x-3)}$ +$\frac{1(x-1)}{(x-1)(x-2)(x-3)}$ =$\frac{2(x-2)}{(x-1)(x-2)(x-3)}$
⇒1(x-3)+1(x-1)-2(x-2)=0
⇔x-3+x-1-2x+4=0
⇔0x=0
Vậy pt có vô số nghiệm
c) ĐKXĐ x$\neq$ 0;x$\neq$ 2
ta có $\frac{x-1}{2x^2-4x}$-$\frac{7}{8x}$ =$\frac{5-x}{4x^2-8x}$ -$\frac{1}{8x-16}$
⇔$\frac{x-1}{2x(x-2)}$ -$\frac{7}{8x}$ =$\frac{5-x}{4x(x-2)}$ -$\frac{1}{8(x-2)}$
⇔$\frac{4(x-1)}{8x(x-2)}$ -$\frac{7(x-2)}{8x(x-2)}$ =$\frac{2(5-x)}{8x(x-2)}$ -$\frac{x}{8x(x-2)}$
⇒4(x-1)-7(x-2)-2(5-x)+x=0
⇔4x-4-7x+14-10+2x+x=0
⇔0x=0
Vậy pt có vô số nghiệm thỏa mãn ĐKXĐ
d)
ta có $\frac{1}{x^2+9x+20}$ +$\frac{1}{x^2+11x+30}$+$\frac{1}{x^2+13x+42}$ =$\frac{1}{18}$
⇔$\frac{1}{(x+4)(x+5)}$ +$\frac{1}{(x+5)(x+6)}$ +$\frac{1}{(x+6)(x+7)}$ =$\frac{1}{18}$
⇔$\frac{18(x+6)(x+7)}{18(x+4)(x+5)(x+6)(x+7)}$ +$\frac{18(x+4)(x+7)}{18(x+4)(x+5)(x+6)(x+7)}$+$\frac{18(x+4)(x+5)}{18(x+4)(x+5(x+6)(x+7)}$ =$\frac{(x+4)(x+5)(x+6)(x+7}{18(x+4)(x+5)(x+6)(x+7)}$
⇒18(x+6)(x+7)+18(x+4)(x+7)+18(x+4)(x+5)-(x+4)(x+5)(x+6)(x+7)=0
⇔18(x+7)(x+6+x+4)+(x+4)(x+5)(18-x^2-13x-42)=0
⇔36(x+7)(x+5)+(x+4)(x+5) (chưa xong)
Bài 19
a) ta có $\frac{2a^2-3a-2}{a^2-4}$ =2
⇔$\frac{(2a-1)(a+2)}{(a-2)(a+2)}$ =2
⇔$\frac{2a-1}{a-2}$ =2
⇔2a-1=2(a-2)
⇔2a-1-2a+4=0
⇔3=0
Vậy không có giá a nào thỏa mãn
b) ĐKXĐ a$\neq$ -1/3;a$\neq$ -3
ta có $\frac{3a-1}{3a+1}$ +$\frac{a-3}{a+3}$ =2
⇔$\frac{(3a-1)(a+3)}{(3a+1)(a+3)}$ +$\frac{(a-3)(3a+1)}{(a+3)(3a+1)}$=$\frac{2(3a+1)(a+3)}{(3a+1)(a+3)}$
⇒(3a-1)(a+3)+(a-3)(3a+1)-2(3a+1)(a+3)=0
⇔ (3a+1)(a+3)(3a-1+a-3-2)=0
⇔(3a+1)(a+3)(4a-6)=0
⇔3a+1=0;a+3=0;4a-6=0
⇔3a+1=0⇒a=-1/3 (KTM)
⇔a+3=0⇒a=-3(KTM)
⇔4a-6=0⇒a=3/2
Vậy a=3/2
c) ĐKXĐ a$\neq$ -3
ta có $\frac{10}{3}$-$\frac{3a-1}{4a+12}$-$\frac{7a+2}{6a+18}$=2
⇔$\frac{40(a+3)}{12(a+3)}$-$\frac{3(3a-10}{12(a+3)}$-$\frac{2(7a+2)}{12(a+3)}$ =$\frac{24(a+3)}{12(a+3)}$
⇒40(a+3)-3(3a-1)-2(7a+2)-24(a+3)=0
⇔40a+120-9a+3-14a-4-24a-72=0
⇔-7a=47
⇔a=-47/7
vậy a=-47/7
d) ĐKXĐ a$\neq$ 5/2;a$\neq$ 2/3
ta có $\frac{2a-9}{2a-5}$+$\frac{3a}{3a-2}$=2
⇔$\frac{(2a-9)(3a-2)}{(2a-5)(3a-2)}$+$\frac{3a(2a-5)}{(3a-2)(2a-5)}$= $\frac{2(3a-2)(2a-5)}{(3a-2)(2a-5)}$
⇒(2a-9)(3a-2)+3a(2a-5)-2(2a-5)(3a-2)=0
⇔(2a-5)(3a-2)(2a-9+3a-2)=0
⇔(2a-5)(3a-2)(5a-11)=0
⇔2a-5=0⇒a=5/2(KTM);3a-2=0⇒a=2/3(KTM);5a-11=0⇒a=11/5
Vậy a=11/5
...................................chúc bạn học tốt..................
