a) \(\left[0,\left(37\right)+0,\left(62\right)\right]\cdot x=10\)

=> \(\left[\frac{37}{99}+\frac{62}{99}\right]\cdot x=10\)

=> \(1\cdot x=10\Rightarrow x=10\)

b) \(\frac{0,\left(12\right)}{1,\left(6\right)}=\frac{\frac{12}{99}}{\frac{5}{3}}=\frac{12}{99}\cdot\frac{3}{5}=\frac{4}{55}\)

=> \(\frac{4}{55}=x:0,\left(4\right)\)

=> \(\frac{4}{55}=x:\frac{4}{9}\)

=> \(x:\frac{4}{9}=\frac{4}{55}\)

=> \(x=\frac{4}{55}\cdot\frac{4}{9}=\frac{16}{495}\)

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

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

B=5/9.2,(37)+0,(5).3,(62)-2 căn bậc hai (-2/3)^2

B=5/9.235/99+5/9.359/99-4/3

B=5/9.(235/99+359/99)-4/3

B=5/9.6-4/3

B=10/3-4/3

B=2

K MK NHA 

Bài 1:

a) \(0,\left(3\right)+3\frac{1}{3}+0,\left(31\right)\)

\(=\frac{1}{3}+\frac{10}{3}+\frac{31}{99}\)

\(=\frac{11}{3}+\frac{31}{99}\)

\(=\frac{394}{99}.\)

b) \(\frac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)

\(=\frac{4}{9}+\frac{1219}{990}-\frac{13}{99}\)

\(=\frac{553}{330}-\frac{13}{99}\)

\(=\frac{139}{90}.\)

Bài 2:

\(0,\left(37\right).x=1\)

\(\Rightarrow\frac{37}{99}.x=1\)

\(\Rightarrow x=1:\frac{37}{99}\)

\(\Rightarrow x=\frac{99}{37}\)

Vậy \(x=\frac{99}{37}.\)

Chúc bạn học tốt!

\(tử:=\dfrac{1}{2}\left[sin\left(60^o-x+30^o-x\right)+sin\left(60^o-x-30^2+x\right)\right]+\dfrac{1}{2}\left[sin\left(30^o-x+60^o-x\right)+sin\left(30^o-x-60^o+x\right)\right]\)

\(=\dfrac{1}{2}\left[2sin\left(\dfrac{\pi}{2}-2x\right)+sin\left(\dfrac{\pi}{6}\right)+sin\left(-\dfrac{\pi}{6}\right)\right]=\dfrac{1}{2}.\left[2sin\left(\dfrac{\pi}{2}-2x\right)+0\right]=sin\left(\dfrac{\pi}{2}-2x\right)=cos2x\)

\(VT=\dfrac{cos2x}{sin4x}=\dfrac{cos2x}{2sin2x.cos2x}=\dfrac{1}{2sin2x}=\dfrac{1}{4sinx.cosx}=\dfrac{\dfrac{1}{cos^2x}}{\dfrac{4sinx.cosx}{cos^2x}}=\dfrac{1+tan^2x}{\dfrac{4sĩnx}{cosx}}=\dfrac{1+tan^2x}{4tanx}=VP\)

\(164:\left\{192-\left[2^3\cdot15-\left(40-37\right)^2\right]+2021^0\right\}\)

\(=164:\left\{192-120+9+1\right\}\)

\(=164:82=2\)

a) (- 16) . (- 7) . 5 = [(- 16) . 5] . (- 7) = 560.

b) 11 . (- 12) + 11 . (- 18) = 11 . [(- 12) + (- 18)] = 11 . [- (12 + 18)] = 11 . (- 30) = - 330.

c) 87 . (- 19) – 37 . (- 19) = (- 19) . (87 – 37) = (- 19) . 50 = - 950.

d) 41 . 81 . (- 451) . 0 = 0.

a) (- 16) . (- 7) . 5 = [(- 16) . 5] . (- 7) = 560.

b) 11 . (- 12) + 11 . (- 18) = 11 . [(- 12) + (- 18)] = 11 . [- (12 + 18)] = 11 . (- 30) = - 330.

c) 87 . (- 19) – 37 . (- 19) = (- 19) . (87 – 37) = (- 19) . 50 = - 950.

d) 41 . 81 . (- 451) . 0 = 0.

a) (-16).(-7).5

=16.7.5  = 2.8.7.5

=(2.5).(8.7) = 56 . 10 =560

b) 11 . (-12) + 11.(-18)

=11 . (-12 -18)

=11.(-20) = -220

c) 87.(-19) -  37 .(-19)

=(-87).19 + 37.19

=19 . (-87+37)

=19.(-50)

=  -950

d) 41.81.(-451).0

= (mọi số nhân 0 đều bằng 0)