Question

1pe 2 + 1pe 3+1pe 6

(5pe 4+0.75):2=

(4+3•6):(5+6)=

3 la puterea 3 +(-2)totu la puterea 2

1 1pe2 +2 1pe 3-3
1 si 2 sunt in fata fractiilor

0,1(6)•12=

(-2,8):(0,7)+12:2=

(1pe 2+1pe3):0,(3)-2,5=

(2√2+√8+√18):√2=

Answer 1

1/2 + 1/3+1/6=(3+2+1)/6=6/6=1(5/4+0.75):2=(5/4+75/100):2=(5x25+75)/100x1/2=(125+75)/200=200/200=1

(4+3•6):(5+6)=(4+18):11=22:11=2

3³ +(-2)²=27+4=31 

1 1pe2 +2 1pe 3-3=3/2+7/3-3=(3x3+7x2-3x6)/6=(9+9-18)/6=0

0,1(6)•12=(16-1)/9x12=15/9x12=simplif.cu 3=15/3x4=5x4=20

(-2,8):(0,7)+12:2=-28/10:7/10+6=-28/10x10/7+6=-4+6=2

(1/2+1/3):0,(3)-2,5=(3+2)/6:3/9-25/10=5/6x9/3-25/10=5/2-25/10=(25-25)/10=0

(2√2+√8+√18):√2=(2√2+2√2+3√2):√2=7√2:√2=7√2/√2=7