Odpowiedź :
a) (x+1)(x-2)+(x+2)(x-1) = x² - 2x + x - 1 = x² - x - 1
b) (5k+n)(k-2n)-k(4k+n) = 5k²-10kn + kn - 2n² - 4k² - kn = k² - 10kn - 2n²
c) (4y+3)(y-2)-2y(4-2y) = 4y² - 8y + 3y - 6 - 8y + 4y² = 8y² - 6 - 13y
d) -(3c-2)(c-1)-c(c-2) = (-3c - 2)(c - 1) - c² + 2c = -3c² + 3c - 2c + 1- c² + 2c = -4c² + 3c +1
e) 2(3a²-1)-(a+2)(a+2) = 6a² - 2 - (a² + 2a + 2a + 4) = 6a² - 2 - a² -2a - 2a - 4 = 5a² - 6 - 4a
f) 5x²(x²+1)-(x²-2)(x²+3) = 5x⁴+ 5x² - x⁴- 3x²+ 2x² - 6 = 4x⁴+ 4x² - 6
2.Zapisz w postcia sumy algebraiczne pole trapuze o podstawach x,x+2 i wysokości x+1.
P = 0,5(x + x + 2)(x+1) = (x+1)(x+1) = x² + x + x + 1 = x² + 2x + 1
b) (5k+n)(k-2n)-k(4k+n) = 5k²-10kn + kn - 2n² - 4k² - kn = k² - 10kn - 2n²
c) (4y+3)(y-2)-2y(4-2y) = 4y² - 8y + 3y - 6 - 8y + 4y² = 8y² - 6 - 13y
d) -(3c-2)(c-1)-c(c-2) = (-3c - 2)(c - 1) - c² + 2c = -3c² + 3c - 2c + 1- c² + 2c = -4c² + 3c +1
e) 2(3a²-1)-(a+2)(a+2) = 6a² - 2 - (a² + 2a + 2a + 4) = 6a² - 2 - a² -2a - 2a - 4 = 5a² - 6 - 4a
f) 5x²(x²+1)-(x²-2)(x²+3) = 5x⁴+ 5x² - x⁴- 3x²+ 2x² - 6 = 4x⁴+ 4x² - 6
2.Zapisz w postcia sumy algebraiczne pole trapuze o podstawach x,x+2 i wysokości x+1.
P = 0,5(x + x + 2)(x+1) = (x+1)(x+1) = x² + x + x + 1 = x² + 2x + 1
1
a)
(x+1)(x-2)+(x+2)(x-1)=(x²+2x+x-2)+(x²-x+2x-2)=x²+2x+x-2+x²-x+2x-2=x*4+4x-4
b)
(5k+n)(k-2n)-k(4k+n)=(5k²-10kn+kn-2n²)-(-4k²-kn)=5k²-10kn+kn-2n²4k²+kn=9k²-12kn
c)
(4y+3)(y-2)-2y(4-2y)=(4y²-8y+3y-6)-(-8y+4y²)=4y²-8y+3y-6+8y-4y²=3y-6
d)
-(3c-2)(c-1)-c(c-2)=-(3c²-3c-2c+2)-(-c²+2c)=-3c²+3c+2c-2+c²-2c=-2c²+c
e)
2(3a²-1)-(a+2)(a+2)=(6a²-2)-(a²+2a+2a+4)=6a²-2-a²-2a-2a-4=4a²-2-4a
f)
5x²(x²+1)-(x²-2)(x²+3)=(5x*4+5x²)-(x*4+3x²-2x²+6)=5x*4+5x²-x*4-3x²+2x²-6=4x*4-6
2
P=1/2(a razy b)h
p=(x+x+2)(x+1)=1/2(2x+2)(x+1)=1/2(2x²+2x+2x+2)=1/2(2x²+4x+2)=1/2 razy 2(x²+2x+1)=x²+2x+2
*-potęga
/-ułamek
łap
a)
(x+1)(x-2)+(x+2)(x-1)=(x²+2x+x-2)+(x²-x+2x-2)=x²+2x+x-2+x²-x+2x-2=x*4+4x-4
b)
(5k+n)(k-2n)-k(4k+n)=(5k²-10kn+kn-2n²)-(-4k²-kn)=5k²-10kn+kn-2n²4k²+kn=9k²-12kn
c)
(4y+3)(y-2)-2y(4-2y)=(4y²-8y+3y-6)-(-8y+4y²)=4y²-8y+3y-6+8y-4y²=3y-6
d)
-(3c-2)(c-1)-c(c-2)=-(3c²-3c-2c+2)-(-c²+2c)=-3c²+3c+2c-2+c²-2c=-2c²+c
e)
2(3a²-1)-(a+2)(a+2)=(6a²-2)-(a²+2a+2a+4)=6a²-2-a²-2a-2a-4=4a²-2-4a
f)
5x²(x²+1)-(x²-2)(x²+3)=(5x*4+5x²)-(x*4+3x²-2x²+6)=5x*4+5x²-x*4-3x²+2x²-6=4x*4-6
2
P=1/2(a razy b)h
p=(x+x+2)(x+1)=1/2(2x+2)(x+1)=1/2(2x²+2x+2x+2)=1/2(2x²+4x+2)=1/2 razy 2(x²+2x+1)=x²+2x+2
*-potęga
/-ułamek
łap
a) (x+1)(x-2)+(x+2)(x-1) = x² - 2x + x - 1 = x² - x - 1
b) (5k+n)(k-2n)-k(4k+n) = 5k²-10kn + kn - 2n² - 4k² - kn = k² - 10kn - 2n²
c) (4y+3)(y-2)-2y(4-2y) = 4y²-8y+3y-6-8y+4y² = 8y²-6-13y
d) -(3c-2)(c-1)-c(c-2) = (-3c-2)(c-1) - c²+2c = -3c²+3c-2c+1-c²+2c = -4c²+3c+1
e) 2(3a²-1)-(a+2)(a+2) = 6a²-2-(a²+2a+2a+4) = 6a²-2-a²-2a-2a-4 = 5a²-6-4a
f) 5x²(x²+1)-(x²-2)(x²+3) = 5x⁴+ 5x²-x⁴-3x²+2x²-6= 4x⁴+ 4x²-6
2.Zapisz w postcia sumy algebraiczne pole trapuze o podstawach x,x+2 i wysokości x+1.
P = 0,5(x+x+2)(x+1) = (x+1)(x+1) = x²+x+x+1 = x²+2x+1
b) (5k+n)(k-2n)-k(4k+n) = 5k²-10kn + kn - 2n² - 4k² - kn = k² - 10kn - 2n²
c) (4y+3)(y-2)-2y(4-2y) = 4y²-8y+3y-6-8y+4y² = 8y²-6-13y
d) -(3c-2)(c-1)-c(c-2) = (-3c-2)(c-1) - c²+2c = -3c²+3c-2c+1-c²+2c = -4c²+3c+1
e) 2(3a²-1)-(a+2)(a+2) = 6a²-2-(a²+2a+2a+4) = 6a²-2-a²-2a-2a-4 = 5a²-6-4a
f) 5x²(x²+1)-(x²-2)(x²+3) = 5x⁴+ 5x²-x⁴-3x²+2x²-6= 4x⁴+ 4x²-6
2.Zapisz w postcia sumy algebraiczne pole trapuze o podstawach x,x+2 i wysokości x+1.
P = 0,5(x+x+2)(x+1) = (x+1)(x+1) = x²+x+x+1 = x²+2x+1