Có: \(\dfrac{x+1}{x-1}\cdot\dfrac{x+1}{x-2}=0\left(x\ne1;x\ne2\right)\)
\(\Rightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
tìm x: part 1 : a,(x^3)^2-(x+1)(x-1)=1 b,(x-2)^2-3(x-2)=0 c,(x+2)(x^2-2x+4)-x(x^2+2)=15 d,(x+1)^2-(x+1)(x-2)=0 e,4x(x-2017)-x+2017=0 f,(x+4)^2-16=0 part 2: a,x^3+27+(x+3)(x-9)=0 b,(2x-1)^2-4x^2+1=0 c,2(x-3)+x^2-3x=0 d,x^2-2x+1=6x-6 e,x^3-9x=0
Tìm x
a) 4x(x + 1) = 8(x + 1)
b) x(x – 1) – 2(1 – x) = 0
c) 5x(x – 2) – (2 – x) = 0
d) 5x(x – 200) – x + 200 = 0
e) x3 + 4x = 0
f) (x + 1) = (x + 1)2
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0 d) x2.(x+1)-x.(x+1)+x.(x-1)=0
b) 9.(2x+1)2-4.(x+1)2=0 e) (x-2)2-(x-2).(x+2)=0
c) x3-6x2+9x=0 g) x4-2x2+1=0
h) 4x2+y2-20x-2y+26=0 i) x2-2x+5+y2-4y=0
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0 d) x2.(x+1)-x.(x+1)+x.(x-1)=0
b) 9.(2x+1)2-4.(x+1)2=0 e) (x-2)2-(x-2).(x+2)=0
Tìm x biết:
1> (x+1)^2-(x+2)^2=3
2> (x-1)(x+1)-(x-3)^2=0
3> (x+1)^3-x^2(x+2)-(x-1)^2=0
4> (x+1)(x^2-x+1)-x^2(x+2)+2(x+3)^2=0
a,x+5/x-1+8/x^2-4x+3=x+1/x-3 b,x-4/x-1-x^2+3/1-x^2+5/x+1=0 c,3x/4-5=3-x/2+5x-1/6 d,(x-2)(x+2)-(x-3)(x+4)-2x+3=0 e,(x-1)^2+2(x+1)=5x+5 g,(x-3)(x+4)x=0
1) Tim x
a) 3x(x-1)+x-1=0
b)2(x+3)-x^2-3x=0
làm tương tự như bài này nè!
x(x-2)+x-2=0
=>x*((x-2)-x-3)=0
=>(x-2)(x+10=0
hoac x-3=0 =>x=3
hoac 5x-1=0 =>x=1 phan 5
vậy x = 1 phần 5 ;x=3
hoac 5x-1=0 => x=1 phan 5
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0
b) 9.(2x+1)2-4.(x+1)2=0
c) x3-6x2+9x=0
d) x2.(x+1)-x.(x+1)+x.(x-1)=0
tìm x bt a) x^3=x^5 b) 4x.(x+1)=(x+1) c) x.(x-1)-2(1-x)=0 d) 2x.(x-2)-(2-x)^2 e) (x-3)^2+3-x=0 f) 5x.(x-2)-(2-x)=0
tìm x
A, (x-1)^2+(x-3)^2-2x^2+1=0
B, (x-1)*(x^2+x+1)-(x+1)*(x^2-x+1)+2x=0
C, (x-5)*(x-5)+(2x+1)^2-3x^2=0
D, (x-1)-9=0
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