Lời giải:
$5,24\times 0,43+2,62\times 0,62+0,39\times 5,24$

$=5,24\times 0,43+5,24\times 0,31+0,39\times 5,24$

$=5,24\times (0,43+0,31+0,39)=5,24\times 1,13=5,9212$

x+5,66=9

x=9-5,66

x=3,34

x+5,66=9

x        =9-5,66

x        =3,34

3,34

TICK CHO MÌNH NHÉ!

=(27,34+5,66)+(3,45+3,55)=33+7=40

\(27,34+3,45+5,66+3,55\)

\(=\left(27,34+5,66\right)+\left(3,45+3,55\right)\)

\(=33+7\)

\(=40\)

Chúc bạn học tốt~~

27,34+3,45+5,66+3,55

=(27,34+5,66)+(3,45+3,55)

=33+7=40

TL

Kết quả 

= 28,16

Hok tốt

Mikko biết

TL:

Kết quả là 

= 28,16 

HT 

( 6,7 + 5,66 - 3.7 +4,34 ).( -76,6 .1,2 +7,66 .12 )

= [( 6,7 -3,7 ) + ( 5,66 + 4,34 )].[ -7,66 . 12 + 7,66 . 12 )

= ( 3 + 10 ).[ 12 . ( -7,66 + 7,66 )]

= 13 . 0

= 0

\(\left(6,7-5,66-3,7+4,34\right)\cdot\left(-76,6\cdot1,2+7,66\cdot12\right)\)

\(=\left[\left(6,7-3,7\right)+\left(5,66+4,34\right)\right]\left[-7,66\cdot12+7,66\cdot12\right]=\left(3\cdot10\right)\cdot0=0\)

(6,7+5,66-3,7+4,34).(-76,6.1,2+7,66.12)

= (6,7 + 5,66 - 3,7 + 4,34) . [7,66 . (12 - 12)]

= (6,7 + 5,66 - 3,7 + 4,34) . 7,66 . 0

= 0

Lời giải:
Gọi $d=ƯCLN(2n+5, 8n+24)$

$\Rightarrow 2n+5\vdots d; 8n+24\vdots d$

$\Rightarrow 8n+24-4(2n+5)\vdots d$

$\Rightarrow 4\vdots d$ (1)

Vì $2n+5\vdots d$, mà $2n+5$ lẻ nên $d$ lẻ (2)

Từ $(1); (2)\Rightarrow d=1$

$\Rightarrow 2n+5, 8n+24$ nguyên tố cùng nhau.

$\Rightarrow BCNN(2n+5, 8n+24)=(2n+5)(8n+24)$

=(6,7+5,66 -3,7+4,34).(-7,66.12+7,66.12)

=(6,7+5,66 -3,7+4,34).[(-7.66+7.66).12]

=(6,7+5,66 -3,7+4,34).(0.12)

=(6,7+5,66 -3,7+4,34).0

=0 

Ta có:

-76,6 .1,2 + 7,66 . 12 

= -7,66 .10 .1,2 + 7,66 . 12

= -7,66 . 12 + 7,66 .12

= 12 . ( -7,66 + 7,66 )

= 12 . 0

= 0

\(\Rightarrow\left(6,7+5,66-3,7+4,34\right).\left(-76,6.1,2+7,66.12\right)=\left(6,7+5,66-3,7+4,34\right).0=0\)

21,6 - X x 3 = 5,24 + 4,76

21,6 - X x 3 = 10

          X x 3 = 21,6 - 10

          X x 3 = 11,6 

          X       = 11,6 : 3

          X       = 58/15

\(21,6-x\times3=5,24+4,76\)

\(21,6-x\times3=10\)

\(x\times3=21,6-10\)

\(x\times3=11,6\)

\(x=11,6:3\)

\(x=\dfrac{58}{15}\)