Từ \(3x=7y\Rightarrow\frac{x}{7}=\frac{y}{3}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có

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

=>x=28

    y=12

\(3x=7y\Rightarrow\frac{x}{3}=\frac{y}{7}\)

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

\(\frac{x}{7}=-4\Rightarrow x=-28\)

\(\frac{y}{3}=-4\Rightarrow y=-12\)

\(3x=7y\) <=> \(\frac{x}{7}=\frac{y}{3}\) và \(x-y=16\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

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

=> \(x=28\)

=> \(y=12\)

ta có 3x=7y=> x/y=7/3=>x/7=y/3=x-y /7-3=-16/4=-4

=> x=-4.7=-28

y=-4.3=-12

bài 2 :

ta có x:y:z=3:5:(-2)

=>x/3=y/5=z/-2

=>5x/15=y/5=3z/-6

áp dụng tc dãy ... ta có :

5x/15=y/5=3z/-6=5x-y+3z/15-5+(-6)=-16/4=-4

=>x/3=-=>x=-12

=>y/5=-4=>y=-20

=>z/-2=-4=>z=8

b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)

Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)

c) \(x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)

b) 3x(x\(^3\) +12x-2x\(^2\)-8=0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
 

a) x\(^2\)+5x=0

x(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
 

c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36

x\(^3\)-27+x(x+2)(2-x)=36

4x-27=36

4x=36+27

4x=63

x=\(\dfrac{63}{4}\)