\(a+\frac{2}{44}=\frac{24}{12}\)

\(a=\frac{24}{12}-\frac{2}{44}\)

\(a=\frac{43}{22}\)

Vậy \(a=\frac{43}{22}\)

\(\frac{a+2}{44}=\frac{24}{12}\Leftrightarrow\left(a+2\right).12=44.24\Rightarrow a+2=\frac{44.24}{12}=\frac{44.2.12}{12}=44.2=88\Rightarrow a=86\)

Ta có:\(a+\frac{2}{44}=\frac{24}{12}\)

    \(a=\frac{24}{12}-\frac{2}{44}\)

   \(a=\frac{43}{22}\)

ta có

a+2/44=24/22

a=24/12-2/24=23/12

a=24/12-2/44

a=2-1/22

a=44/22-1/22

a=43/22

Phương pháp giải:

- Muốn tìm một số hạng ta lấy tổng trừ đi số hạng kia.

- Muốn tìm số bị trừ ta lấy hiệu cộng với số trừ.

Lời giải chi tiết:

a) x + 6 = 24

    x = 24 − 6

    x = 18

b) x − 12 = 44

    x = 44 + 12

    x = 56

D

D

D

a: \(12=2^2\cdot3;18=3^2\cdot2\)

=>\(ƯCLN\left(12;18\right)=2\cdot3=6\)

b: \(24=2^3\cdot3;48=2^4\cdot3\)

=>\(ƯCLN\left(24;48\right)=2^3\cdot3=24\)

c: \(44=2^2\cdot11;121=11^2\)

=>\(ƯCLN\left(44;121\right)=11\)

d: \(36=3^2\cdot2^2;108=3^3\cdot2^2;224=2^5\cdot7\)

=>\(ƯCLN\left(36;108;224\right)=2^2=4\)

a, Ta có :

\(12=2\cdot2\cdot3=2^2\cdot3\)

\(18=2\cdot3\cdot3=3^2\cdot2\)

Thừa số nguyên tố chung là : \(2;3\)

\(ƯCLN\) \(\left(12;18\right)=2\cdot3=6\)

⇒ \(ƯCLN\) \(\left(12;18\right)\) là \(6\)

b, Ta có:

\(24=2\cdot2\cdot2\cdot3=2^3\cdot3\)

\(48=2\cdot2\cdot2\cdot2\cdot3=2^4\cdot3\)

Thừa số nguyên tố chung là : \(2;3\)

\(ƯCLN\) \(\left(24;48\right)\) \(=2^3\cdot3=8\cdot3=24\)

⇒ \(ƯCLN\) \(\left(24;48\right)\) là \(24\)

c, Ta có

\(44=2\cdot2\cdot11=2^2\cdot11\)

\(121=11\cdot11=11^2\)

Thừa số nguyên tố chung là : \(11\)

\(ƯCLN\) \(\left(44;121\right)\) \(=11\)

⇒ \(ƯCLN\) là \(11\)

d, Ta có :

\(36=2\cdot2\cdot3\cdot3=2^2\cdot3^3\)

\(108=2\cdot2\cdot3\cdot3\cdot3=2^2\cdot3^3\)

\(224=2\cdot2\cdot2\cdot2\cdot2\cdot7=2^5\cdot7\)

Thừa số nguyên tố chung là : \(2\)

\(ƯCLN\) \(\left(36;108;224\right)\) \(=2\cdot2=2^2=4\)

⇒ \(ƯCLN\) \(\left(36;108;224\right)\) là \(4\)

\(#thaolinh\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(24 \div a+24=30\)

`=> 24 \div a = 30 - 24`

`=> 24 \div a = 6`

`=> a = 24 \div 6`

`=> a = 4`

Vậy, `a=4`

`b)`

\( 1974 \div 14+a \times 141=423\)

`=> 141 + a \times 141 = 423`

`=> a \times 141 = 423 - 141`

`=> a \times 141 = 282`

`=> a = 282 \div 141`

`=> a = 2`

Vậy, `a=2`

`c)`

\(120-(45-a \times 4)=115\)

`=> 45 - a \times 4 = 120 - 115`

`=> 45 - a \times 4 = 5`

`=> a \times 4 = 45 - 5`

`=> a \times 4 = 40`

`=> a = 40 \div 4`

`=> a = 10`

Vậy, `a=10`

`d)`

\((a \times 4-44) \div 4=44\)

`=> a \times 4 - 44 = 44 \times 4`

`=> a \times 4 - 44 = 176`

`=> a \times 4 = 176 + 44`

`=> a \times 4 = 220`

`=> a = 220 \div 4`

`=> a = 55`

Vậy, `a = 55.`

a) 24:a+24=30

 24:a= 30 - 24

24:a=6

a= 24:6

a=4

______

b) 1974:14+ax141=423 

141 + a x 141 = 423

141 x (a+1) = 423

a+1 = 423 : 141

a+1=3

a=3-1

a=2

_____
c) 120-(45-ax4)=115

45 - a x 4= 120 - 115

45 - a x 4= 5

a x 4 = 45 - 5

a x 4 = 40

a = 40:4

a=10

______

  d) (ax4-44):4=44

a x 4 - 44 = 44  x 4

a x 4 - 44 = 176

a x 4 = 176 + 44

a x 4 = 220

a= 220:4

a=55

a) \(24:a+24=30\)

\(\Rightarrow24:a=30-24\)

\(\Rightarrow24:a=6\)

\(\Rightarrow a=24:6\)

\(\Rightarrow a=4\)

b) \(1974:14+a\cdot141=423\)

\(\Rightarrow141+a\cdot141=423\)

\(\Rightarrow141\cdot\left(1+a\right)=423\)

\(\Rightarrow1+a=423:141\)

\(\Rightarrow1+a=3\)

\(\Rightarrow a=3-1\)

\(\Rightarrow a=2\)

c) \(120-\left(45-a\cdot4\right)=115\)

\(\Rightarrow45-a\cdot4=120-115\)

\(\Rightarrow45-a\cdot4=5\)

\(\Rightarrow a\cdot4=45-5\)

\(\Rightarrow a=40:4\)

\(\Rightarrow a=10\)

d) \(\left(a\cdot4-44\right):4=44\)

\(\Rightarrow a\cdot4-44=44\cdot4\)

\(\Rightarrow a\cdot4-44=176\)

\(\Rightarrow a\cdot4=176+44\)

\(\Rightarrow a\cdot4=220\)

\(\Rightarrow a=220:4\)

\(\Rightarrow a=55\)

a) 2 × 9 + 63 = 18 + 63 = 81

b) 24 : 4 + 44 = 6 + 44 = 50

c) 5 × 6 – 12 = 30 – 12 = 18

d) 3 × 7 – 16 = 21 – 16 = 5