Question

Bài 1: Tìm x

a, (8-5x)(x+2)+4(x-2)(x+1)+(x-2)(x+2)=0

b, (8x-3)(3x+2)-(4x+7)(x+4)=(4x+1)(5x-1)-33

Bài 2: Cm các đẳng thức sau:

a, (x+y)(x³-x²y+xy^2-y^3)=x^4+y^4

b (a-1)(a-2)+(a-3)(a+4)-(2a^2+5a-34)=24-7a

c. (a+c)(a-c)-b(2a-b)-(a-v+c)(a-b-c)=o

Answer 1

a) <=> (8-5x+x-2)(x+2) + 4(x^2-x-2)=0

<=> 6x +12 - 4x^2 - 8x +4x^2 -4x -8 =0

<=> -6x -4 = 0

<=> x= 4/6

 

Answer 2

Ta có VT =\(a^2-c^2-2ab+b^2-\left[\left(a-b\right)^2-c^2\right]\)

= \(a^2-c^2-2ab+b^2-\left(a^2-2ab+b^2\right)+c^2\)

=\(a^2-c^2-2ab+b^2-a^2+2ab-b^2+c^2\)

= 0 =VP (đpcm)