a) 0,(37)+0,(62) = 1

Có 0.(37)=\(\frac{37}{99}\)và  0.(62) = \(\frac{62}{99}\)

  \(\frac{37}{99}\)+     \(\frac{62}{99}\)= 1

\(\Rightarrow0,\left(37\right)+0.\left(62\right)=1\)

b)\(0,\left(37\right)\times3=1\)

Có: \(0,\left(37\right)=\frac{37}{99}\)

       \(\frac{37}{99}\times3=1\)

    \(\Rightarrow0\left(37\right)\times3=1\)

\(0,\left(37\right)+0,\left(62\right)=0,\left(99\right)\) 

Theo quy ước làm tròn số ta dược :

\(0,\left(99\right)\approx1\) (đpcm)

b) Làm tương tự câu a) ta có :

\(0,\left(33\right).3=0,\left(99\right)\approx1\) (đpcm)

đề câu a sai r` = 11 ms đúng 

ta có:0,(37)=1/99*37=37/99

0,(62)=1/99*62=62/99

0,(37)+0,(62)=37/99+62/99=99/99=1(đpcm)

\(0,\left(37\right)+0,\left(62\right)=\frac{37}{99}+\frac{62}{99}=\frac{99}{99}=1\)

\(0,\left(33\right).3=\frac{33}{99}.3=\frac{1}{3}.3=\frac{3}{3}=1\)

a, 0,(37)+0,(62)=0,(99) 
Theo quy ước làm tròn số ta dược :
0,\left(99\right)\approx10,(99)≈1 (đpcm)
b) Làm tương tự câu a) ta có :
0,\left(33\right).3=0,\left(99\right)\approx10,(33).3=0,(99)≈1 (đpcm)

a) Ta có:
0,\left(37\right)=\frac{37}{99}0,(37)=9937​ ; 0,\left(62\right)=\frac{62}{99}0,(62)=9962​
=> 0,\left(37\right)+0,\left(62\right)=\frac{37}{99}+\frac{62}{99}=\frac{99}{99}=10,(37)+0,(62)=9937​+9962​=9999​=1
b) Ta có:
0,\left(33\right)=\frac{33}{99}0,(33)=9933​
=> 0,\left(33\right).3=\frac{33}{99}.3=\frac{1}{3}.3=10,(33).3=9933​.3=31​.3=1

\(0,\left(37\right)+0,\left(62\right)=\dfrac{37}{99}+\dfrac{62}{99}=\dfrac{99}{99}=1\)

\(0,\left(33\right)\cdot3=\dfrac{1}{3}\cdot3=1\)

a, 0,(37)+0,(62)=1

ta có : 0,(37)=37/99

           0,(62)=62/99

=> 0,(37)+0,(62)=37/99+62/99=99/99=1

Vậy 0,(37)+0,(62)=1

b, 0,(33).3=1

ta có : 0,(33)=33/99=1/3

=> 0,(33).3=1/3.3=1

Vậy 0,(33).3=1

0,(37)+0,(62)=0,(99) 
Theo quy ước làm tròn số ta dược :
0,\left(99\right)\approx10,(99)≈1 (đpcm)
b) Làm tương tự câu a) ta có :
0,\left(33\right).3=0,\left(99\right)\approx10,(33).3=0,(99)≈1 (đpcm)

con chó

a)ta có: 0, (37) + 0, (62) = 1

\(\Rightarrow\)\(\dfrac{37}{99}+\dfrac{62}{99}=1\left(ĐPCM\right)\)

b)ta có: 0, (33).3=1

\(\Rightarrow\)\(\dfrac{1}{3}.3=1\left(ĐPCM\right)\)

a) Ta có:

0, (37) = 0, (01) . 37 = \(\dfrac{1}{99}\) . 37 = \(\dfrac{37}{99}\)

0, (62) = 0, (01) . 62 = \(\dfrac{1}{99}\) . 62 = \(\dfrac{62}{99}\)

\(\Rightarrow\)0, (37) + 0, (62) = \(\dfrac{37}{99}\) + \(\dfrac{62}{99}\) = \(\dfrac{99}{99}\)= 1

Vậy 0, (37) + 0, (62) = 1 (ĐPCM)

b) Ta có:

0, (33) = 0, (01) . 33 = \(\dfrac{1}{99}\) . 33 = \(\dfrac{33}{99}\)

\(\Rightarrow\)0, (33) . 3 = \(\dfrac{33}{99}\) . 3 =\(\dfrac{99}{99}\) = 1

Vậy 0, (33) . 3 = 1 (ĐPCM)

tick mk nhé