Question

1){111-[400:(a-5)+260]:3}×9+32=239

2)121-{90:5-4×[20-2(3×a+2)]}=119

3)Daca x=18 si xy+xz=1224,atunci y+z este ?

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Answer 1

1)
{111-[400:(a-5)+260]:3}×9+32=239
{111-[400:(a-5)+260]:3}×9=239-32
{111-[400:(a-5)+260]:3}×9=207
111-[400:(a-5)+260]:3=207:9
111-[400:(a-5)+260]:3=23
[400:(a-5)+260]:3=111-23
[400:(a-5)+260]:3=88
400:(a-5)+260=88×3
400:(a-5)+260=264
400:(a-5)=264-260
400:(a-5)=4
a-5=400:4
a-5=100
a=100+5
a=105


2)
121-{90:5-4×[20-2×(3×a+2)]}=119
18-4×[20-2×(3×a+2)]=121-119
18-4×[20-2×(3×a+2)]=2
4×[20-2×(3×a+2)]=18-2
4×[20-2×(3×a+2)]=16
20-2×(3×a+2)=16:4
20-2×(3×a+2)=4
2×(3×a+2)=20-4
2×(3×a+2)=16
3×a+2=16:2
3×a+2=8
3×a=8-2
3×a=6
a=6:3
a=2


3)Daca x=18 si xy+xz=1224,atunci y+z este ?

x=18
xy+xz=1 224
înlocuim....
18y + 18z = 1 224
18×(y+z)=1 224
y+z=1 224 : 18
y+z=68