Rozwiązane

10-2×(3⅕-2¼×⅓)=


(1200×600)÷10⁴-(2+7×100÷(2×5))=


(1-7/9)×(5,2-0.7)+(1,5-1⅓)×4⅘=
7/9 ułamek zwykły



Odpowiedź :

10-2×(3⅕-2¼×⅓)=
10-2×(¹⁶/₅-⁹/₄×⅓)=
10-2×(¹⁶/₅-¾)=
10-³²/₅+³/₂=
10-6,4+1,5=
3,6+1,5=
5,1

(1200×600)÷10⁴-(2+7×100÷(2×5))=
(72000):10⁴-(2+700:10)=
7,2×10⁴:10⁴-(2+70)=
7,2-72=-64,8

(1-7/9)×(5,2-0,7)+(1,5-1⅓)×4⅘=
(2/9)×4,5+(3/2-4/3)×24/5=
(2/9)×9/2+(9/6-8/6)×24/5=
1+1/6×24/5=1+24/30=1 i 12/15
Annaa300
10-2×(3⅕-2¼×⅓)=10-2×(3⅕-9/4x1/3)=
10-2×(3⅕-1/2)= 10-2×(3 i 2/10-5/10)=
10-2x(2 12/10 – 5/10)= 10-2x 2 i7/10=
10-2x 27/10=10 -27/5= 10 – 5 i 2/5=
9 i 5/5 – 5 i2/5= 4 i3/5


(1200×600)÷10⁴-(2+7×100÷(2×5))=120000:10000-(2+7×100÷(2×5))=
12 –(2+ 700:10)=12 –(2+70)=12 -72= -60

(1-7/9)×(5,2-0.7)+(1,5-1⅓)×4⅘=
2/9 x 4,5 +(1i ½ - 1 i 1/3)x 4 i 4/5=
2/9 x 45/10 +(1 i3/6- 1i 2/6) x 24/5=
1+(1i 1/6)x 24/5=
1+7/6 x 24/5=1 +28/5=
1+5 i 3/5=6 i 3/5

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