x*0,75+2.4/3.8-0.8=6.4

x*0.75+2.4/3=6.4

x*0.75+2.4/3=19.2/3

gach mau so ta co

x*0.75+2.4=19.2

x*0.75=19.2-2.4

x*0.75=16.8

x=16.8/0.75

x=22.4

\(\frac{x\times0,75+2,4}{3,8-0,8}\)

\(=\frac{x\times0,75+2,4}{3}\)

\(x\times0,75=3-2,4\)

\(x\times0,75=0,6\)

\(x=0,6\div0,75\)

\(x=0,8\)

tao ko biết

mày ngu như con bò ấy trần anh duân ạ . mày ngu hơn cả con bò bằng 22.4

k bn nha

a: \(\Leftrightarrow2x=\dfrac{19}{5}:\dfrac{3}{32}=\dfrac{608}{15}\)

hay x=304/15

b: \(\Leftrightarrow0.25x=20\)

hay x=80

c: \(\Leftrightarrow x=\dfrac{1}{100}:\dfrac{5}{2}=\dfrac{2}{500}=\dfrac{1}{250}\)

d: \(\Leftrightarrow0.1x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)

hay \(x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{20}{5}=4\)

Bài 1 :

Áp dụng t.c tỉ lệ thức là làm dc thôi bn

Bài 2 :

Đặt :

\(\dfrac{x}{4}=\dfrac{y}{7}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)\(\left(1\right)\)

Thay \(xy=112\) vào \(\left(1\right)\) ta có :

\(4k.7k=112\)

\(\Leftrightarrow28k^2=112\)

\(\Leftrightarrow k^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

+) \(k=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4.2=8\\y=7.2=14\end{matrix}\right.\)

+) \(k=-2\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4.\left(-2\right)=-8\\y=7.\left(-2\right)=-14\end{matrix}\right.\)

Vậy ..

\(\frac{15^3.8^2}{5^6.4^4}=\frac{\left(3.5\right)^3.\left(2^3\right)^2}{5^6.\left(2^2\right)^4}=\frac{3^3.5^3.2^6}{5^6.2^8}=\frac{3^3}{5^3.2^2}=\frac{27}{500}\)

\(=3\cdot2^3\cdot5^2+2\cdot2^6\cdot2^2\cdot3+2^3\cdot2\cdot3\cdot2^2+3\cdot2\cdot3\cdot2^2\\ =2^3\cdot3\cdot5^2+2^9\cdot3+2^6\cdot3+2^3\cdot3^2\\ =2^3\cdot3\left(5^2+2^6+2^3+3\right)=24\left(25+64+8+3\right)\\ =24\cdot100=2400\)