ai lước qua mà ko ticks thì sau này cha mẹ người ko ticks sẽ chết bất đắc kì tử
TH1: x-3=6x+3 TH2: x-3=-6x-3
<=> x-3-6x-3=0 <=>x-3+6x+3=0
<=>-6-5x=0 <=>7x=0
<=>x=6/5 <=> x=0
Vậy x ={0;6/5}
Bài 3: phân tích thành nhân tử:
1/ 9x^3-xy^2
2/x^2-3xy-6x+18y
3/x^2-3xy-6x+18y 3/6x(x-y)-9y^2+9xy
4/ 6xy-x^2+36-9y^2
5/ x^4-6x^2+5
6/ 9x62-6x-y^2+2y
Bài 4:Tìm x, biết:
1/ (x-1)(x^2+x+1)-x^3-6x=11
2/ 16x^2-(3x-4)^2=0
3/ x^3-x^2+3-3x=0
4/ x-1/x+2=x+2/x+1
5/1/x+2/x+1=0
6/ 9-x^2/x : (x-3)=1
Bài5: 1/ 12x^3y^2/18xy^5
2/10xy-5x^2/2x^2-8y^2
3/ x^2-xy-x+y/x^2+xy-x-y
4/ (x+1)(x^2-2x+1)/(6x^2-6)(x^3-1)
5/ 2x^2-7x+3/1-4x^2
x-3/x+3 - x+3/x-3= 6x^2/9-6x^2
Bài 1. Thu gọn:
a) x2 – 4 – (x + 2)2 | b) (x + 2)(x – 2) – (x – 3)(x + 1) |
c) (x – 2)(x + 2) – (x – 2)(x + 5) | d) (6x + 1)2 + (6x – 1)2 – 2(6x + 1)(6x – 1) |
e) 7a(3a – 5) + (2a -3)(4a + 1) – (6a – 2)2 | g) (5y – 3)(5y + 3) – (5y – 4)2 |
h) (3x + 1)3 – (1 – 2x)3 | i) (2x + 1)2 + 2(4x2 – 1) + (2x – 1)2 |
4x(5x − 2) 7x Ä 3x 2 − 6x + 2ä b) c) 2x(3x + 2) + (4x + 3)(2x − 1) 3x 3 y 2 : x 2 d) Ä x 3 + 4x 3 − 6x 2 ä : 4x 2 e) Ä 3x 2 − 6x ä f) : (2 − x) Ä 6x 2 + 13x − 5 ä g) : (2x + 5) Ä x 3 − 3x 2 + x − 3 ä h) : (x − 3)
Rút gọn:
A=(x^4 - 3x^2 + 9)(x^2 + 3) + (3 - x^2)^2
M= 5(x+2y)^2 - (3y + 2x)^2 + (4x - y)^2 + 3(x - 2y)(x + 2y)
E= (6x + 1)^2 + (6x - 1)^2 - 2(1 + 6x)(6x - 1)
C= (x^2 + 4)(x + 2)(x -2) - (x^2 - 3)^3
thu gọn
a) 5y.(2y-1)-(3y+2).(3-3y)
b) (6x+1)^2 -2(6x+1).(6x-1)+(6x-1)^2
c) (2x+3)-2(2x+3)(x-2)+(x-2)^2
1, (x-2)(x+2)(x^2+4)-(x^2-3)(x^2+3)
2, (6x+1)^2 +(6x-1)^2 -2(1+6x)(6x-1)
giúp mik với
1) (y-z)(12x^2-6x)+(y-z)(12x^2+6x)
2) a(b-c)+d(b-c)-c(c-b)
3) a(b-3)+(3-b)-b(3-b)
4) 5(a-b)^2+(a+b)(a-b)
5) 2a(x-y)-(y-x)
6) a^2(x-y)-(x-y)
7) 6x^2 -3+7x(6x^2-3)+4y(3-6x^2)
Tìm x Toán đại 8 Hằng đẳng thức đáng nhớ?
Tìm x:
1. (x-1)^3+3.(x-3)^2-(x+2).(x^2-2x+4) = (x+2)^3-(x-3).(x^2+9)-6x^2+5
2.(3+2x)^3-(6x-1).(6x+1) = (2x-1)^3+(x+4)^2-x^3+(x+1).(x^2+x+1)
phân tích đa thức
x^4+6x^3+11x^2+6x+1
x^4+x^3+x^2+x+1
6x^4+5x^3-38x^2+5x+6
x^4+5x^3-12x^2+5x+1
