xy=-30

yz=42

Do đó: \(\frac{xy}{yz}=\frac{-30}{42}=\frac{-5}{7}\)

=>\(\frac{x}{z}=-\frac57\)

=>z=-1,4x

z-x=-12

=>-1,4x-x=-12

=>-2,4x=-12

=>x=5

=>\(z=-1,4\cdot5=-7\)

xy=-30

=>5y=-30

=>y=-6

Giúp tôi giải toán ! tick nhé!

Giúp tôi giải toán tick nhé!

đây

suốt ngày hỏi

Đặt ba đơn thức lần lượt là a,b,c

ta có:a*b*c= (-1/2019.x^4.y.z^3).(108.x^3.y^2.z).(x^5.y.z^4)

d=(-1/2019.108.304).(x^4.x^3.x^5.y.y^2.y.z^3.z.z^4)

d=-32832.x^12.y^4.z^8

=> d<0 với mọi x,y,z do x^12.y^4.z^8 luôn dương

=> đpcm

\(x-y=-30\Rightarrow\dfrac{x}{-30}=\dfrac{1}{y}\\ y.z=-42\\ \Rightarrow\dfrac{z}{-42}=\dfrac{1}{y}\\ \Rightarrow\dfrac{x}{-30}=\dfrac{z}{-42}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{-30}=\dfrac{z}{-42}=\dfrac{z-x}{-42-\left(-30\right)}=\dfrac{-12}{-12}=1\)

\(\dfrac{x}{-30}=1\Rightarrow x=-30\\ \dfrac{z}{-42}=1\Rightarrow z=-42\)

\(x.y=-30\Rightarrow-30.y=-30\Rightarrow y=1\)

46452007

\(\left\{{}\begin{matrix}x,y,z\ne0\\x^2.y.z=-4\\xy^2z=2\\xyz^2=-2\end{matrix}\right.\)\(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\\\left(4\right)\end{matrix}\)

(2).(3).(4) \(\left(x^2yz\right).\left(xy^2z\right)\left(xyz^2\right)=\left(x^{2+1+1}.y^{1+2+1}.z^{1+1+2}\right)=\left(xyz\right)^4=\left(-4\right).2.\left(-2\right)=8\)\(\Leftrightarrow\left[{}\begin{matrix}xyz=2\\xyz=-2\end{matrix}\right.\)\(\begin{matrix}\left(I\right)\\\left(II\right)\end{matrix}\)

TH(I)

(2) => x =-2 ;(3) => y =1;(4) => z =-1

TH(II)

(2) => x =2 ; (3) => y =-1; (4) => z =1

(x;y;z)=(-2;1;-1);(2;-1;1)