Question

Calculați:

a)

(3x+2)²+(x-5)(x+5)

b)

(4x-1)²-(x-3)(x+3)

c)

(5-4x)(4x+5)+(4x-3)²

d)

(7x+2)(2-7x)+(6x-1)²

e)

(√2x-1)(√2x+1)-(7x-3)²

f)

(4x+5)²-(√5x+4)(4-√5x)

Trebuie cu rezolvare!!!

Answer 1

(3x+2)²+(x-5)(x+5)=9x^2+12x+4+x^2-25=10x^2+12x-21
(4x-1)²-(x-3)(x+3)=16x^2-8x+1-x^2+9=15x^2-8x+10
(5-4x)(4x+5)+(4x-3)²=25-16x^2+16x^2-24x+9=-24x+34
(7x+2)(2-7x)+(6x-1)²=4-49x^2+36x^2-24x+1=-13x^2-24x+5
Daca si x se afla sub radical in urmatoarele cazuri, ai:
(√2x-1)(√2x+1)-(7x-3)²=2x-1-49x^2+42x-9=-49x^2+44x-10
(4x+5)²-(√5x+4)(4-√5x)=16x^2+40x+25-16+5x=16x^2+45x+9
Daca la ultimele doua x e in afara radicalului:
(√2x-1)(√2x+1)-(7x-3)²=2x^2-1-49x^2+42x-9=-47x^2+42x-10
(4x+5)²-(√5x+4)(4-√5x)=16x^2+40x+25-16+5x^2=21x^2+40x+9
PS: Uita-te exact la exercitii ca sa stii ce varianta scrii.

Answer 2

a)

(3x+2)²+(x-5)(x+5)

[tex]\it (3x+2)^2 = (3x)^2 +2\cdot3x\cdot2+2^2 = 9x^2 +12x+4 \\ \\ (x-5)(x+5) = x^2 - 5^2 = x^2 - 25 \\ \\ \\ (3x+2)^2 +(x-5)(x-5) = 9x^2+12x+4+x^2-25 = \\ \\ =10x^2+12x-21 [/tex]

c)

(5-4x)(4x+5)+(4x-3)²


[tex]\it (5-4x)(4x+5)=(5-4x)(5+4x)= 5^2-(4x)^2 = 25 - 16x^2 \\ \\ (4x-3)^2 = (4x)^2-2\cdot4x\cdot3+3^2=16x^2-24x+9[/tex]


f)

(4x+5)²-(√5x+4)(4-√5x)

[tex]\it (4x + 5)^2 = (4x)^2 +2\cdot4x\cdot5 + 5^2 = 16x^2+40x + 25 \\ \\ (\sqrt5+4)(4-\sqrt5 x) = (4+\sqrt5x)(4-\sqrt5 x) = 4^2-(\sqrt5 x)^2 = \\ \\ = 16 - 5x^2[/tex]

[tex]\it (4x+5)^2 - (\sqrt5x+4)(4-\sqrt5x) = 16x^2+40x+25-(16 - 5x^2 ) = \\ \\ = 16x^2+40x+25- 16 + 5x^2 = 21x^2 + 40x + 9[/tex]