Đáp án:
Giải thích các bước giải:
d/(3x+2)/2-(3x+1)/6=2x+5/3
=>3(3x+2)/6-(3x+1)/6=12x/6+10/6
=>3(3x+2)-(3x+1)=12x+10
=>9x+6-3x-1=12x+10
=>9x-3x-12x=10-6+1
=>-6x=3
=>x=-1/2
e/ x-(2x-2)/5+(x+8)/6=7+(x-1)/3
=>30/x-6(2x-2)/30+5(x+8)/30=210/30+10(x-1)/30
=>30-6(2x+2)+5(x+8)=210+10(x-1)
=>30-12x-12+5x+40=210+10x-10
=>-12x+5x-10x=210-10-30+12-40
=>-17x=142
=>x=-142/17
f/ 2(x-3)/7-x+2=(13x+4)/21
=> 6(x-3)/21-21x/21+42/21=(13x+4)/21
=>6(x-3)-21x+42=13x+4
=>6x-18-21x+42=13x+4
=>6x-21x-13x=4+18-42
=>-28x=-20
=>x=5/7
Bài 2:
a/ 2x(x-3)+5(x-3)=0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>2x=-5 =>x=3
=>x=-2,5
b/x^2 -4 -( x -2 ) ( 3-2x) = 0
<=>x^2-4-(x-2)(3-2x)=0
<=>x^2 - 4 - 3x + 2x^2 - 6 + 4x = 0
<=> 3x^2 -7x+2= 0
<=> x=1/3 và x=2
c/(2x+5)^2=(x+2)^2
=>(2x+5-x-2)(2x+5+x+2)=0
=>(x-3)(3x+7)=0
=>x-3=0 hoặc 3x+7=0
=>x=3 =>3x=-7
=>x=-7/3
d/ x^2-5x+6=0
=>x^2-2x+3x+6=0
=>x(x-2)+3(x+2)=0
=>x(x-2)=0 hoặc 3(x+2)=0
=> x-2=0 =>x+3=0
=>x=-3
e/ 2x^3+6x^2=x^2+3x
=>2x^2(x+3)-x(x+3)=0
=>x(2x-1)(x+3)=0
=>x(2x-1)=0 hoặc x+3=0
=>2x-1=0 =>x=-3
=>2x=1
=>x=1/2
f/ (x+1/2)^2+2(x+1/x)-8=0
bài cuối cùng mk ko bik nhé!
=>x=2
