Rozwiązane

Przekształć iloczyny na sumy:
a) (a-b)(a+b)=
b) (2x-1)(2x+1)=
c) (4x+1)(4x-1)=
d) (5y+2z)(5y-2x)=
e) (3x-4y)(4y+3x)=
f) (4a+7b)(7b-4a)=
g) ( ⅖x+3)(⅖x-3)=
h) (0.3a-2)(2+0,3a)=
i) (2,5+3)(2,5-3)=
j) (7,2x+!)
k) (5x+½y)(5x-½y)
l) ( 1½x-y)(1,5+y)=
m) (3/7a+⅔b)(⅔-3/7a)=
n) ( 11x-0,5z)(11x+0,5z)=
o) (1⅓z+⅘a)(1⅓z-0,8a)=
p) (10a=8b)(8b+10a)=



Odpowiedź :

Tewuka
a) (a-b)(a+b)= a²-b²
b) (2x-1)(2x+1)=4x²-1²
c) (4x+1)(4x-1)=16x²-1²
d) (5y+2x)(5y-2x)=25y²-4x²
e) (3x-4y)(4y+3x)=(3x-4y)(3x+4y)=9x²-16x²
f) (4a+7b)(7b-4a)=(7b+4a)(7b-4a)= 49b²-16a²
g) ( ⅖x+3)(⅖x-3)=0,16x²-9
h) (0.3a-2)(2+0,3a)=(0,3a-2)(0,3a+2)=0,09a²-4
i) (2,5+3)(2,5-3)=6,25-9
j) (7,2x+!)= 51,84x²-!²
k) (5x+½y)(5x-½y)=25x²-¼y²
l) ( 1½x-y)(1,5+y)=2,25-y²
m) (3/7a+⅔b)(⅔-3/7a)=4/9-13,69a²
n) ( 11x-0,5z)(11x+0,5z)=121x²-0,25z²
o) (1⅓z+⅘a)(1⅓z-0,8a)=1i 7/9z² - 0.64a²
p) (10a-8b)(8b+10a)=100a²-64b²
Annaa300
a) (a-b)(a+b)= a²-b²
b) (2x-1)(2x+1)=4x²-1
c) (4x+1)(4x-1)=16x²-1
d) (5y+2x)(5y-2x)=25y²-4x²
e) (3x-4y)(4y+3x)=(3x-4y)(3x+4y)=9x²-16x²
f) (4a+7b)(7b-4a)=(7b+4a)(7b-4a)= 49b²-16a²
g) ( ⅖x+3)(⅖x-3)=4/25x²-9
h) (0.3a-2)(2+0,3a)=(0,3a-2)(0,3a+2)=0,09a²-4
i) (2,5x+3)(2,5x-3)=6,25x-9
j) (7,2x+1)²= 51,84x²-1
k) (5x+½y)(5x-½y)=25x²-¼y²
l) ( 1½x-y)(1,5x+y)=2,25x²-y²
m) (3/7a+⅔b)(⅔-3/7a)=4/9-13,69a²
n) ( 11x-0,5z)(11x+0,5z)=121x²-0,25z²
o) (1⅓z+⅘a)(1⅓z-0,8a)=16/9z² - 0.64a²
p) (10a-8b)(8b+10a)=100a²-64b²

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