Question

Mă poate ajuta cineva vă rog frumos? Ex 2,4​

Answer 1

Răspuns:

Explicație pas cu pas:

2) (2x-y)² = 4x²-4xy+y²

(4+2x)² = 16+16x+4x²

(2√2-3a)(2√2+3a) = 4·2-3a·3a = 8-9a²

(x/3+2y)³ = (x/3)³+(2y)³+3·(x/3)·2y·(x/3+2y) =

= x³/27 + 8y³ +2xy(x/3 +2y) = x³/27 + 8y³+2x²y/3 + 4xy²

(y/2 -4z)³ = (y/2)³ - (4z)³ - 3·y/2 ·4z·(y/2 -4z) =

= y³/8 - 64z³ - (12yz/2)·(y/2-4z) =

= y³/8 -64z³ - 3y²z + 24yz²

(3x-2y+√5)² = (3x)²+(2y)²+(√5)²-2·3x·2y-2·2y·√5+2·3x·√5 =

= 9x²+4y²+5-12xy-4y√5+6x√5

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

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

c) (x+√2)²-8 = (x+√2)² - (2√2)² = (x+√2-2√2)(x+√2+2√2) =

= (x-√2)(x+3√2)

d) (2x+√3)²-27 = (2x+√3)²-(3√3)² = (2x+√3-3√3)(2x+√3+3√3) =

= (2x-2√3)(2x+4√3) = 4(x-√3)(x+2√3)

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

= (-1-3√2x)(5√2x-1) = (1+3√2x)(1-5√2x)

f) a⁶-b⁶ = (a³-b³)(a³+b³) = (a-b)(a²+ab+b²)(a³+b³)

g) x²-2√3x+3 - (y-2√3)² = (x-√3)² -  (y-2√3)² =

= (x-√3-y+2√3)(x-√3+y-2√3) =

= (x-y+√3)(x+y-3√3)

h) 14a - 49a² + 100b²-1 = -[(7a-1)² - 100b²] =

= -[(7a-1)² - (10b)²] =-(7a-1-10b)(7a-1+10b) =

= (10b-7a+1)(10b+7a-1)