Matematică, întrebare adresată de Miha2004, 9 ani în urmă

1.000-(432:4+278x2):4x5+714:7a=612:9

5+[16-(9:ax8+7)]x6=4:(1x2)+(3x3)

10x{a-10x[362+10x(24+14:4)]}=99+10:10

315:(a-8)=5

277:a=3 rest.1

[ax9+(111-202:20)]x2=160

Răspunsuri la întrebare

Răspuns de CarMina03
1
1.000-(432:4+278x2):4x5+714:7a=612:9
1.000-(108+556):4x5+714:7a=68
1.000-664:4x5+714:7a=68
1.000-166x5+714:7a=68
1.000-830+714:7a=68
170+714:7a=68
714:7a=68-170

714:7a=--102

7a=-714:102
7a=-7
a=-1


5+[16-(9:ax8+7)]x6=4:(1x2)+(3x3)

5+[16-(9:ax8+7)]x6=4:2+9


5+[16-(9:ax8+7)]x6=2+9


5+[16-(9:ax8+7)]x6=11


[16-(9:ax8+7)]x6=11-5


[16-(9:ax8+7)]x6=6

16-(9:ax8+7)=6:6

16-(9:ax8+7)=1

9:ax8+7=16-1

9:ax8+7=15

9:ax8=15-7

9:ax8=8

9:a=8:8
9:a=1
a=9x1
a=9


10x{a-10x[362+10x(24+14:4)]}=99+10:10

10x{a-10x[362+10x(24+3,5)]}=99+1

10x{a-10x[362+10x27,5]}=100

{a-10x[362+275]}=100:10

a-10x637=10

a-6370=10

a=10+6370

a=6380


315:(a-8)=5

(a-8)=315:5

a-8=63

a=63+8

a=71


277:a=3 rest.1

277=ax3+1

277-1=ax3

276=ax3

a=276:3

a=92


[ax9+(111-202:2)]x2=160


[ax9+(111-101)]x2=160

[ax9+10]x2=160

ax9+110=160:2

ax9+10=80

ax9=80-10

ax9=70

a=70:9

a=7,(7)




Miha2004: Mersi ult :*
Miha2004: La ultima era impartit la 2 am gresit eu :((
CarMina03: ok,imediat
Miha2004: Ms :*
Miha2004: ..
Răspuns de 0000000
1
[tex]1000-(432:4+278*2):4*5+714:7a=612:9 \\ 1000-(108+556):4*5+102a=68 \\ 664:4*5+102a=1000-68 \\ 166*5+102a=932 \\ 830+102a=932 \\ 102a=932-830 \\ 102a=102 \\ \boxed{a=1} \\ \\ 5+[16-(9:a*8+7)]*6=4:(1*2)+(3*3) \\ 5+[16-(72:a+7)]*6=4:2+9 \\ 5+[16-(72:a+7)]*6=11 \\ 6[16-(72:a+7)]=6 \\ 16-(72:a+7)=1 \\ 72:a+7=15 \\ 72:a=8 \\ \boxed{a=9} \\ \\ 10*{a-10*[362+10*(24+14:4)]}=99+10:10 \\ 10*{a-10*[362+10(24+3,5)]}=100 \\ 10*[a-10*(362+10*27,5)]=100 \\ a-10*(362+275)=100:10 \\ a-10*637=10 \\ a-6370=10 \\ \boxed{a=6380}[/tex]

[tex]315:(a-8)=5 \\ a-8=315:5 \\ a-8=63 \\ a=63+8 \\ \boxed{a=71} \\ \\ 277:a=3~rest~1 \\ 277=3a+1 \\ 3a=277-1 \\ 3a=276 \\ \boxed{a=92} \\ \\ \([9a+(111-202:2)]*2=160 \\ 9a+(111-101)=160:2 \\ 9a+10=80 \\ 9a=80-10 \\ 9a=70 \\ \boxed{a=7,(7)}[/tex]

Miha2004: Ms mult :*
Alte întrebări interesante