x- 32:16=48

=>x-2=48

=>x=48+2

=>x=50

(x-32):16=48

=>x-32= 48x 16

=>x-32= 768

=>x=768+32

=>x=800

1.x=50

2.x=bấm máy tính nhá mik hơi bí chút

k ủng hộ nhá

1. x-32:16=48

         x-32=48.16

        x-32=768

            x=768+32

             x=800

Vậy x=800

2.(x-32):16=48

Tương tự như câu trên nha bn

*** mk nha

\(\dfrac{x}{3}-\dfrac{x^2}{4}=0\\ \Leftrightarrow x\left(\dfrac{1}{3}-\dfrac{x}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(\dfrac{x}{3}-\dfrac{x^2}{4}=0\)

\(\Leftrightarrow\dfrac{4x-3x^2}{12}=0\)

\(\Leftrightarrow-3x^2+4x=0\)

\(\Leftrightarrow-3x\left(x-\dfrac{4}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)

0.0625

k nha

bằng 1 phần 16 đấy

mình chắc như vậy

<=>\(\frac{24-18-X}{6}\)=\(\frac{3X}{6}\)

<=>\(6-X=3X\)

<=>\(-4X=-6\)

<=>\(X=\frac{3}{2}\)

x-y = 5 => x= 5+y 

\(\frac{x-4}{y-3}=\frac{3}{4}\Rightarrow4x-16=3y-9\Leftrightarrow4x-3y=7\)

Thay x= 5+y vào biểu thức ở đề bài: \(4\left(5+y\right)-3y=7\Leftrightarrow20+4y-3y=7\Leftrightarrow y=-13\)

=> x= 5+y = -8

\(x+x:0,5+x:0,25+x:0,125=2022\)

\(x\text{×}1+x\text{×}2+x\text{×}4+x\text{×8}=2022\)

\(x\text{×}\left(1+2+4+8\right)=2022\)

\(x\text{×}15=2022\)

\(x=2022:15\)

\(x=134,8\)

\(x^3=\frac{27}{8}=\left(\frac{3}{2}\right)^3=>x=\frac{3}{2}\)