1.+2.LGR
( 2x + 41 y )
2
Binomische Formeln LÖSUNG
= 4x2 + xy +
1
16
(3y
y2
(0,2x + 1)2= 0,04x2 + 0,4x + 1
( 2m − 32 n)
2
= 4m2 –
8
9
mn +
4
9
(0,5l – 1)2 = 0,25l2 – l + 1
2
+
1z
2
2
) = 9y
4
COOL 7
+ 3y2z +
1
4
z2
(4a3 + 2b2)2 = 16a6 + 16a3b2 + 4b4
n2
( 4x
2
−
2
3
)
2
= 16x4 –
16
3
x² +
4
9
(5m2 – 8n3)2 = 25m4 – 80m2n3 + 64n6
( 52 + k ) . ( 52 − k ) = 254 – k
(9x − 31 ) . (9x + 31 ) = 81x – 91
2
2
(1,2r – s2) . (1,2r + s2) = 1,44r2 – s4
(13x3 – 2y2) . (13x3 + 2y2) = 169x6 – 4y4
Ergänze fehlende Stellen!
(12 + b) . (12 – b) = 144 – b2
(5k2 – 4)2 = 25k2 – 40k2 + 16
(2x + 4y)2 = 4x2 + 16xy + 16y2
( 31 x 2 – 9) . ( 31 x 2 + 9) =
(8a – 1)2 = 64a2 – 16a + 1
(1,3z2 – 1) . (1,3z2 + 1) = 1,69z4 – 1
(l + 10) . (l – 10) = l2 – 100
(x2 + y3)2 = x4 + 2x2y3 + y6
1.+2.LGR
1 x4
9
– 81
Gleichnamige Bruchterme addieren & subtrahieren
LÖSUNG
COOL 7
1.+2.LGR
Herausheben – Abfahrt oder Slalom LÖSUNG
COOL 7
a) 4u + 4v = 4 . (u + v)
b) 2,3r – 2,3s = 2,3(r – s)
a) rx – ry = r . (x – y)
b) kl + km = k . (l + m)
a) k5 – k4 = k4 . (k – 1)
b) a2b – a2 = a2 . (b – 1)
a) 6x2y3 + 18xy5 = 6xy3 . (x + 3y2)
b) 25v4w2 – 15v3w = 5v3w . (5vw – 3)
a) 4e3 + 12e2 – 8e = 4e . (e2 + 3e – 2)
b) 18q3 + 12q2 – 6 = 6 . (3q3 + 2q2 – 1)
a) 36k3 + 27k2 – 18k = 9k . (4k2 + 3k – 2)
b) 21f3 – 14f2 + 7f = 7f . (3f2 – 2f + 1)
Hebe (–1) heraus!
a) –a + b = (–1) . (a – b)
b) x – y = (–1) . (–x + y)
a) –2x – 2y – 2z = (–1) . (2x + 2y + 2z)
b) –2a + 3b – 4c = (–1) . (2a – 3b + 4c)
Ergänze, was fehlt!
a) 3ab + 3c = 3 . (ab + c)
b) 12 – 6x = 6 . (2 – x)
a) –x + xy = –x . (1 – y)
b) –ab – 6a = –a . (b + 6)
Vereinfache durch Herausheben und Kürzen!
a)
2x 3 − 3x
2x
=
a)
5x − 10y
15x + 5y
=
(
x. 2x 2 − 3
2x
5. ( x − 2y )
5. ( 3x + y )
)
=
=
2x 2 − 3
2
b)
x − 2y
3x + y
b)
4a2 − 4a
8a2
6z5 − 3z2
2
9z + 6z
3
=
=
4a. ( a − 1)
8a2
(
a −1
= 2a
)
3z2 2z3 − 1
2z3 − 1
= 3 + 2z
3z ( 3 + 2z )
2
Hebe den Klammerausdruck heraus!
a) 3g . (3h – i) – (3h – i) = (3h – i) . (3g – 1)
b) (5k + 2) – (5k + 2) . r = (5k + 2) . (1 – r)
a) (c – x) . (3k – m) – (c – x) . (m – 4k) =
(c – x) . [(3k – m) – (m – 4k)]
b) (4z – 1) . (x + 2y) – z . (x + 2y) =
(x + 2y) . [(4z – 1) – z]
aus:GOLLMANN U.A., Lebendige Mathematik 4, öbv&hpt Wien, 2005, S.63
1.LGR
Doppelbrüche LÖSUNG
COOL 7
aus:GOLLMANN U.A., Lebendige Mathematik 4, öbv&hpt Wien, 2005, S.63
2.LGR
Doppelbrüche LÖSUNG
COOL 7
3.LGR
Binomische Formeln LÖSUNG
(x + 2)2 = x2 + 4x +4
(z + 1)2 = z2 + 2z +1
(3a + b)2 = 9a2 + 6ab +b2
(4u + 3v)2 = 16u2 +24uv + 9v2
(a – 3)2 = a2 – 6a + 9
(2k – 1)2 = 4k2 – 4k +1
(4y – z)2 = 16y2 – 8yx + z2
(5k – 3l)2 = 25k2 – 30kl + 9l2
(x – 4) . (x + 4) = x4 – 16
(7m + 1) . (7m – 1) = 49m2 – 1
(6u – v) . (6u + v) = 36u2 – v2
(4k – 5l) . (4k + 5l) = 16k2 – 25l2
COOL 7
Ergänze fehlende Stellen!
(12 + b) . (12 – b) = 144 – b2
(2x + 4y)2 = 4x2 + 16xy + 16y2
aus:GOLLMANN U.A., Lebendige Mathematik 4, öbv&hpt Wien, 2005, S.54ff
3.LGR
Gleichnamige Bruchterme addieren & subtrahieren
COOL 7
LÖSUNG
aus:GOLLMANN U.A., Lebendige Mathematik 4, öbv&hpt Wien, 2005, S.63
3.LGR
Doppelbrüche LÖSUNG
COOL 7
© Veritas Linz
3.LGR
Herausheben – Abfahrt oder Slalom LÖSUNG
COOL 7
a) 4u + 4v = 4 . (u + v)
b) 2,3r – 2,3s = 2,3(r – s)
a) rx – ry = r . (x – y)
b) kl + km = k . (l + m)
a) k5 – k4 = k4 . (k – 1)
b) a2b – a2 = a2 . (b – 1)
a) 6x2y3 + 18xy5 = 6xy3 . (x + 3y2)
b) 25v4w2 – 15v3w = 5v3w . (5vw – 3)
a) 4e3 + 12e2 – 8e = 4e . (e2 + 3e – 2)
b) 18q3 + 12q2 – 6 = 6 . (3q3 + 2q2 – 1)
a) 36k3 + 27k2 – 18k = 9k . (4k2 + 3k – 2)
b) 21f3 – 14f2 + 7f = 7f . (3f2 – 2f + 1)
Ergänze, was fehlt!
a) 3ab + 3c = 3 . (ab + c)
b) 12 – 6x = 6 . (2 – x)
a) –x + xy = –x . (1 – y)
b) –ab – 6a = –a . (b + 6)
Hebe den Klammerausdruck heraus!
a) 3g . (3h – i) – (3h – i) = (3h – i) . (3g – 1)
b) (5k + 2) – (5k + 2) . r = (5k + 2) . (1 – r)
a) (c – x) . (3k – m) – (c – x) . (m – 4k) =
(c – x) . [(3k – m) – (m – 4k)]
b) (4z – 1) . (x + 2y) – z . (x + 2y) =
(x + 2y) . [(4z – 1) – z]

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