a) 4x(x + 1) = 8(x + 1) g) 5x(x – 2000) – x + 2000 = 0 b) x(x – 1) – 2(1 – x) = 0 h) x2 – 4x = 0 c) 2x(x – 2) – (2 – x)2 = 0 k) (1 – x)2 – 1 + x = 0 d) (x – 3)3 + 3 – x = 0 m) x + 6×2 = 0 e) 5x(x – 2) – (2 – x) = 0 n) (x + 1) = (x + 1)2
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a) 4x(x + 1) = 8(x + 1) g) 5x(x – 2000) – x + 2000 = 0 b) x(x – 1) – 2(1 – x) = 0 h
By Nevaeh
Đáp án:
a) 4x(x + 1) = 8(x + 1)
4x(x + 1)+8(x + 1)=0
(x + 1).(4x+8)=0
TH1:x+1=0
x =0-1
x=-1
TH2:4x+8=0
4x =0-8
4x =-8
x =-8:4
x=-2
Vậy x∈{-1;-2}
b) x(x – 1) – 2(1 – x) = 0
x(x – 1) + 2(x – 1) = 0
(x – 1).(x+2)=0
TH1:x – 1=0
x =0+1
x =1
TH2:x+2=0
x =0-2
x =-2
Vậy x ∈{1;-2}
c) 2x(x – 2) – (2 – x)2 = 0
2x(x – 2) +(x – 2)2 =0
(x – 2).(2x+2)=0
TH1:x-2=0
x =0+2
x =2
TH2:2x+2=0
2x =0-2
2x =-2
x =-2:2
x =-1
Vậy x∈{2;-1}
d) (x – 3)3 + 3 – x = 0
(x – 3)3 – (x-3)=0
(x – 3).(3-1)=0
TH1: x-3=0
x =0+3
x =3
TH2:3-1=0(vô lí)
Vậy x =3
e) 5x(x – 2) – (2 – x) = 0
(x – 2).(5x-1)=0
TH1:x-2=0
x =0+2
x =2
TH2:5x-1=0
5x =0+1
5x =1
x =1:5
x =1/5
Vậy x∈{2;1/5}
g) 5x(x – 2000) – x + 2000 = 0
5x(x – 2000) –( x – 2000)=0
(x – 2000).(5x-1)=0
TH1: x-2000=0
x =0+2000
x =2000
TH2:5x-1=0
5x =0+1
5x =1
x =1:5
x =1/5
Vậy x∈{2000;1/5}
h) x2 – 4x = 0
x.(2-4)=0
x.(-2)=0(vô lí)
vậy x ∈∅
k) (1 – x)2 – 1 + x = 0
(1 – x)2 –( 1 – x)=0
(1 – x).(2-1)=0
TH1:1-x=0
x =1-0
x =1
TH2:2-1=0(vô lí)
vậy x =1
m) x + 6×2 = 0
x+ 12x =0
x.(1+12) =0
x.13 =0(vô lí)
vậy x ∈∅
n) (x + 1) = (x + 1)2
(x + 1) +(x + 1)2=0
(x + 1).(1+2)=0
(x + 1).3=0
x=-1
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