II gimnazjum

Policzmy to razem 2

Matematyka

a) x2+6x+9=x2+2⋅x⋅3+32=(x+3)2

a) m2−10m+25=m2−2⋅m⋅5+52=(m−5)2

a   - bok troˊjkąta T2​ (w cm) a+4   - bok troˊjkąta T1​ (w cm)   Roˊz˙nica poˊl tych troˊjkątoˊw:

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a) A+B=(x−1)+(x2+x)= &nbsp; x−1+x2+x= &nbsp;    =x2+2x−1 &nbsp; &nbsp; b) A⋅B=(x−1)⋅(x2+x)=

n, n+1, n+2, n+3, n+4   - pięcˊ kolejnych liczb naturalnych

Zwróć uwagę, że każdą liczbę czterocyfrową postaci abcd można zapisać jako 1000a+100b+10c+d (np. 4567=4∙1000+5∙100+6∙10+7) &nbsp; Nasza liczba czterocyfrowa:&nbsp; 1000a+100b+10c+d &nbsp; Liczba powstała po przestawieniu cyfr:&nbsp; 1000d+100c+10b+a &nbsp; Liczba powstała po dodaniu:&nbsp; 1000a+100b+10c+d+1000d+100c+10b+a= =1001a+110b+110c+1001d= =11⋅91a+11⋅10b+11⋅10c+11⋅91d= =11​​(91a+10b+10c+91d)

Skoro ten trójkąt jest prostokątny o przyprostokątnych a, b oraz przeciwprostokątnej c, to z twierdzenia Pitagorasa możemy zapisać:  a2+b2=c2     Dodatkowo pole wynosi 6, więc:  21​ab=6   ∣⋅2   ab=12     Uproścmy najpierw wyrażenie korzystając ze wzorów skróconego mnożenia (mnożymy stopniowo pierwsze dwa nawiasy i ostatnie dwa nawiasy):    (a+b+c)(a+b−c)=[(a+b)+c]⋅[(a+b)−c]=

f(x)=0 31​x−9=0   ∣⋅3 x−27=0   ∣+27

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