\(a,12\)

Vậy số \(12\) khi phân tích ra thừa số nguyên tố là : \(2^2\cdot3\)

\(b,20\)

Vậy số \(20\) khi phân tích ra thừa số nguyên tố là : \(2^2\cdot5\)

\(c,24\)

Vậy số \(24\) khi phân tích ra thừa số nguyên tố là : \(2^3\cdot3\)

\(d,100\)

Vậy số \(100\) khi phân tích ra thừa số nguyên tố là:\(2^2\cdot5^2\)

\(e,120\)

Vậy số \(120\) khi phân tích ra thừa số nguyên tố là: \(2^3\cdot3\cdot5\)

\(f,150\)

Vậy số \(150\) khi phân tích ra thừa số nguyên tố là: \(2\cdot3\cdot5^2\)

một tấm j cơ vậy e?

\(\dfrac{4}{5}+\dfrac{1}{10}\times\dfrac{9}{2}\)

\(=\dfrac{4}{5}+\dfrac{9}{20}\)

\(=\dfrac{16}{20}+\dfrac{9}{20}\)

\(=\dfrac{25}{20}=\dfrac{5}{4}\)

\(a,119+x=20\)

\(\Rightarrow x=20-119\)

\(\Rightarrow x=-99\)

\(b,\left(235-x\right)-81=54\)

\(\Rightarrow235-x=54+81\)

\(\Rightarrow235-x=135\)

\(\Rightarrow x=235-135\)

\(\Rightarrow x=100\)

\(c,135:x+8=13\)

\(\Rightarrow135:x=13-8\)

\(\Rightarrow135:x=5\)

\(\Rightarrow x=27\)

hello newbie >3

=> 11/10;9/100;75/100;248/1000

1,1 ; 0,09 ; 0,75 ; 0,248

`2,`

\(a,\left(0,5\right)^2+2x=\left(0,8\right)^2\)

\(\Rightarrow0,25+2x=0,64\)

\(\Rightarrow2x=0,64-0,25\)

\(\Rightarrow2x=0,39\)

\(\Rightarrow x=0,39:2\)

\(\Rightarrow x=0,195\)

Vậy \(x=0,195\)

\(b,x-\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}\right)=\dfrac{1}{7}-\dfrac{1}{3}\)

\(\Rightarrow x-\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{1}{7}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{7}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}\)

\(\Rightarrow x=\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\dfrac{1}{5}\)

\(\Rightarrow x=0+0+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}\)

Vậy \(x=\dfrac{1}{5}\)

\(c,2x-\left(\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{9}\right)=\dfrac{1}{9}-\dfrac{1}{4}\)

\(\Rightarrow2x-\dfrac{1}{4}-\dfrac{3}{5}+\dfrac{1}{9}=\dfrac{1}{9}-\dfrac{1}{4}\)

\(\Rightarrow2x=\dfrac{1}{9}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{9}\)

\(\Rightarrow2x=\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\dfrac{3}{5}\)

\(\Rightarrow2x=0+0+\dfrac{3}{5}\)

\(\Rightarrow2x=\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{3}{5}:2\)

\(\Rightarrow x=\dfrac{3}{10}\)

Vậy \(x=\dfrac{3}{10}\)