Bài 6:

\(21,251+6,058+0,749+1,042\)

\(=\left(21,251+0,749\right)+\left(6,058+1,042\right)\)

\(=22+7,1\)

\(=29,1\)

___________________

\(1,53+5,309+12,47+5,691\)

\(=\left(1,53+12,47\right)+\left(5,309+5,691\right)\)

\(=14+11\)

\(=25\)

5:

a: =>x/17=5/17

=>x=5

b; =>6+x=7/11*33=21

=>x=15

c: \(\dfrac{12+x}{43-x}=\dfrac{2}{3}\)

=>3x+36=86-2x

=>5x=50

=>x=10

d: \(\dfrac{x}{5}< \dfrac{3}{7}\)

=>x<3/7*5

=>x<15/7

f: 15/26+x/16=46/52

=>x/16=23/26-15/26=8/26=4/13

=>x=4/13*16=64/13

12 + x / 43 - x có tổng là 12 + 43 = 55

Tổng số phần bằng nhau :

 2 + 3 = 5 ( phần )

 Giá trị 1 phần :

 55 : 5 = 11

x là :

  ( 11 x 2 ) - 12 = 10 

b ) x/5 < 3/7 <=> 7x/35  < 15/35

=> 7x = 14

       x = 2

c ) 1 < 11/x < 2

     11/5 > 2 

=> 5 < x < 11

x { 6 ; 7 ; 8 ; 9 ; 10 }

a: 17-3x=11

=>3x=17-11=6

=>\(x=\frac63=2\)

b: Sửa đề: \(120-2\left(x+3\right)=2^2\cdot5^2\)

=>\(120-2\left(x+3\right)=4\cdot25=100\)

=>2(x+3)=120-100=20

=>x+3=10

=>x=10-3=7

c: 11⋮x-2

=>x-2∈{1;-1;11;-11}

=>x∈{3;1;13;-9}

mà x>0

nên x∈{1;3;13}

d: Ta có: \(12=2^2\cdot3;18=2\cdot3^2\)

Do đó: ƯCLN(12;18)=2*3=6

12⋮x; 18⋮x

=>x∈ƯC(12;18)

=>x∈Ư(6)

=>x∈{1;2;3;6}

\(\frac{12+x}{43-x}=\frac{2}{3}\)

\(\Leftrightarrow2\left(43-x\right)=3\left(12+x\right)\)

\(\Leftrightarrow86-2x=36+3x\)

\(\Leftrightarrow86-36=3x+2x\)

\(\Leftrightarrow50=5x\Leftrightarrow x=10\)