Question

Descompuneti in factori :

a) (3x+1)²- (x+3)²
b) (3x+5)²-(x+1)²
c) (x+√2)²-8
d) (2x+√3)²-27
e) (√2x-1)²-32x²
f) a²-b²

VA ROG AJUTATI-MA !! CU REZOLVAREA COMPLETA VA ROG ! MERSI ! MACAR CATEVA .

Answer 1

 (3x+1)²- (x+3)²=(3x+1-(x+3))*(3x+1+(x+3))=(3x+1-x-3)*(3x+1+x+3)=(2x-2)(4x+4)=2(x-1)*4(x+1)=2*4(x-1)(x+1)=8(x-1)(x+1)

(3x+5)²-(x+1)²=(3x+5-(x+1))*(3x+5+(x+1))=(3x+5-x-1)(3x+5+x+1)=(2x+4)(4x+6)=2(x+2)*2(2x+3)=2*2(x+2)(2x+3)=4(x+2)(2x+3)

(x+√2)²-8=(x+√2-√8)*(x+√2+√8)=(x+√2-2√2)*(x+√2+2√2)=(x-√2)(x+3√2)

(2x+√3)²-27=(2x+√3-√27)(2√3+√27)=(2x+√3-3√3)(2x+√3+3√3)=(2x-2√3)(2x+4√3)=2(x-√3)*2(x+2√3)=4(x-√3)(x+2√3)

 (√2x-1)²-32x²=(√2√x-1)²-32x²=(√2x^1/2-1)²-32x²=(√2√x-1-4√2x)(√2√x-1+4√2x)=(√2x-1-4√2x)(√2x-1+4√2x)


a²-b²=(a-b)(a+b)

Answer 2

Formula: (a+b)²=a²+2×a×b+b²
(a-b)²=a²+2×a×b-b²

a) (3x+1)²=(3x)²+2×3x×1+1²
=9x²+6x+1

(x+3)²=x²+2×x×3+3²
=x²+6x+9

9x²+6x+1-x²+6x+9=
=8x²+10
(6x se taie pt ca da 0 si se calculeaza restu)

b) (3x+5)²=9x²+30x+25
(x+1)²=x²+2x+1

9x²+30x+25-x²+2x+1=
=8x²+32x+26

c) (x+√2)²=x²+2×x×√2+(√2)²
=x²+2√2x+2 (radicalu se taie cu ² si ramane 2)
=x²+2√2x+2-8
=x²+2√2x-6

d) (aici se face la fel)-(2x)²+2×2x×√3+(√3)²
=4x²+4√3x+3
4x²+4√3x+3-27=
=4x²+4√3x-24

e) (√2x-1)²=4x²+2√2x-1
4x²+2√2x-1-32x²=
=28x²+2√2x-1

f) a²-b²=(a+b)(a-b)

Sper ca te-am ajutat!!