Жақшаларды ашу. Коэффициент.

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6th Grade

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Жақшаларды ашу. Коэффициент.

6.2.1.5

жақшаны ашу ережелерін білу;

6.2.1.6

коэффициент, ұқсас мүшелер ұғымдарының анықтамаларын білу;

Ортақ көбейткішті жақша сыртына шығару

Кез келген a, b және c рационал сандары үшін көбейтудің үлестірімділік қасиеті орындалады:

(a + b)c = ac + bc.

1-мысал. (–5 + 7) ⋅ (–2) = 2 ⋅ (–2) = –4;

2-мысал. Жақшаларды аш:

–4(–3a + 7); 7a(–5b + 3 – 2m).

Шешуі.

–4(–3a + 7) = –4 ⋅ (–3a) + (–4) ⋅ 7 = 12a + (–28) = 12a – 28. Көбейткіштің таңбасын ескере отырып, жақшаны ашамыз: –4(–3a + 7) = 4 ⋅ 3a – 4 ⋅ 7 = 12a – 28;

7a(–5b + 3 – 2m) = 7a ⋅ (–5b) + 7a ⋅ 3 + 7a(–2m) = –35ab + 21a – 14am,

Multiple Choice

(4,36−6,57)+(−7,36−1,43)=?

-10

-11

-12

-13

Multiple Choice

2,57−5,34)−(7,57+9,66)=?

-18

-19

-20

-21

Multiple Choice

Коэффициентті тап:

−4⋅17c

-68

-50

Multiple Choice

Коэффициентті тап:

−3⋅6x

-3

-18

-6

Multiple Choice

Коэффициентті тап:

−6⋅1,5y

-6

-9

Multiple Choice

Коэффициентті тап: -2a⋅(−0,3b)

0,6

-0,6

-6

Multiple Choice

Коэффициентті тап:

−x⋅(−y)⋅(−4p)⋅(−t)

-4

-1

Multiple Choice

0,3(x−5y−2)−0,8(2y−3x)=?

2,7x-3,1y+0,6

2,7x-3,1y-0,6

2,7x+3,1y-0,6

-2,7x-3,1y-0,6

Multiple Choice

−6(3x+1)−(x−8)=15

13/19

-13/19

11/19

-11/19

Multiple Choice

(3а – 2,4) *(– 5)=?

– 15 + 12a

– 15а -12

15а + 12

– 15а + 12

Үй жұмысы: 670

Multiple Choice

−6(3x+1)−(x−8)=15

13/19

-13/19

11/19

-11/19

Жақшаларды ашу. Коэффициент.

6.2.1.5

жақшаны ашу ережелерін білу;

6.2.1.6

коэффициент, ұқсас мүшелер ұғымдарының анықтамаларын білу;

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