bài 2: giải các pt sau
a, 5x-4/2=16+1/7
b, 3(x-11)/4=3(x+1)/5-2(2x-5)/10
c, x+2/3 + 3(2x-1)/4-5x-3/6= x+5/12
d, 1/2(x+1) + 1/4(x+3)=3-1/3(x+2)
bài 4: giải các phương trình sau
a, x+5/3x-6 – 1/2 =2x-3/2x-4
b, 3/2x-16 + 3x-20/x-8 + 1/8=13x-102/3x-24
c, 1/x+1 – 5/x-2= 15/(x+1) (2-x)
d, 12/1-9x^2= 1-3x/1+3x – 1+3x/1-3x
e, x+1/x-2 – 5/x+2=12/x^2-4+1
f, 2x+1/2x-1 – 2x-1/2x+1= 8/4x^2-1
2.
a) (5x – 4)/2 = (16x + 1)/7
<=> 7(5x-4)/14 = 2(16x+1)/14
<=> 7(5x-4) = 2(16x+1)
<=> 35x – 28 = 32x + 2
<=> 35x – 32x = 2 + 28
<=> 3x = 30
<=> x = 10
Vậy S = { 10}
b)3(x-11)/4 = 3(x+1)/5 – 2(2x-5)/10
<=> 15(x-11)/20 = 12(x+1)/20 – 4(2x-5)/20
<=> 15(x-11) = 12(x+1) – 4(2x-5)
<=> 15x – 165 = 12x + 12 – 8x + 20
<=> 15x + 8x = 12 + 20 – 165
<=> 23x = – 132
<=> x = -109
Vậy S = { -109}
c) x + 2/3 + 3 (2x-1)/4 – 5x-3/6 = x + 5/12
<=> 12x/12 + 8/12 + 9(2x-1)/12 – 2(5x-3)/12 = 12x/12 + 5/12
<=> 12x + 8 + 9(2x-1) – 2(5x-3) = 12x + 5
<=> 12x + 8 + 18x – 9 – 10x + 6 = 12x + 5
<=> 12x – 12x + 18x – 10x = 5 -6 – 8 + 9 – 6
<=> 8x = -6
<=> x = -6/8 = -3/4
Vậy S = { -3/4}
d) (1/2)(x+1) + (1/4)(x+3) = 3 – (1/3)(x+2)
<=> x+1/2 + x+3 / 4 = 3 – x+2 / 3
<=> 6(x+1)/12 + 3(x+3)/12 = 42/12 – 4(x+2)/12
<=> 6(x+1) + 3(x+3) = 42 – 4(x+2)
<=> 6x + 6 + 3x + 9 = 42 – 4x -8
<=> 9x + 4x = -8 – 15
<=> 13x = -23
<=> x = -23/13
Vậy S = { -23/13}
Đáp án:
2.
a) (5x – 4)/2 = (16x + 1)/7
<=> 7(5x-4)/14 = 2(16x+1)/14
<=> 7(5x-4) = 2(16x+1)
<=> 35x – 28 = 32x + 2
<=> 35x – 32x = 2 + 28
<=> 3x = 30
<=> x = 10
Vậy S = { 10}
b)3(x-11)/4 = 3(x+1)/5 – 2(2x-5)/10
<=> 15(x-11)/20 = 12(x+1)/20 – 4(2x-5)/20
<=> 15(x-11) = 12(x+1) – 4(2x-5)
<=> 15x – 165 = 12x + 12 – 8x + 20
<=> 15x + 8x = 12 + 20 – 165
<=> 23x = – 132
<=> x = -109
Vậy S = { -109}
c) x + 2/3 + 3 (2x-1)/4 – 5x-3/6 = x + 5/12
<=> 12x/12 + 8/12 + 9(2x-1)/12 – 2(5x-3)/12 = 12x/12 + 5/12
<=> 12x + 8 + 9(2x-1) – 2(5x-3) = 12x + 5
<=> 12x + 8 + 18x – 9 – 10x + 6 = 12x + 5
<=> 12x – 12x + 18x – 10x = 5 -6 – 8 + 9 – 6
<=> 8x = -6
<=> x = -6/8 = -3/4
Vậy S = { -3/4}
d) (1/2)(x+1) + (1/4)(x+3) = 3 – (1/3)(x+2)
<=> x+1/2 + x+3 / 4 = 3 – x+2 / 3
<=> 6(x+1)/12 + 3(x+3)/12 = 42/12 – 4(x+2)/12
<=> 6(x+1) + 3(x+3) = 42 – 4(x+2)
<=> 6x + 6 + 3x + 9 = 42 – 4x -8
<=> 9x + 4x = -8 – 15
<=> 13x = -23
<=> x = -23/13
Vậy S = { -23/13}
4.
a)x+53x−6−12=2x−32x−4x+53x−6−12=2x−32x−4
<=> x+53(x−2)−12=2x−32(x−3)x+53(x−2)−12=2x−32(x−3)
<=> 2(x + 5) – 3(x – 2) = 3(2x – 3)
<=> 2x + 10 – 3x + 6 = 6x – 9
<=> -x + 16 = 6x – 9
<=> -x = 6x – 9 – 16
<=> -x = 6x – 25
<=> -x – 6x = -25
<=> -7x = -25
<=> x = 25/7
Vậy S={25/7}
b) 3/2x-16 + 3x-20/x-8 + 1/8=13x-102/3x-24 ( ĐKXĐ : x khác 8 )
<=> 3/2x-16 + 3x-20/x-8 – 13x-102/3x-24 = – 1/8
<=> 3/ 2(x – 8) + 3x – 20/ x – 8 – 13x – 102/ 3(x – 8) = – 1/8
<=> 9 + 6( 3x – 20) – 2(13x – 102)/ 6( x – 8) = – 1/8
<=> 93 – 8x / 6 ( x- 8) = – 1/8
<=> 8 ( 93 – 8x) = -6 ( x – 8)
<=> 744 – 64x = -6x + 48
<=> – 64x + 6x = 48 – 744
<=> – 58x = -696
<=> x = 12 ( TM)
Vậy S = {12}
c)ĐKXĐ: x khác -1 và 2
Ta có:
1/(x+1) -5/(x-2) = 15/(x+1)(x-2)
<=> x-2 – 5(x+1) = 15
<=> x = -11/2 (TM ĐKXĐ)
Vậy S={-11/2}