a) = 4x²y³

b) = 4/2x²y

c) = xyz²

- \(5xyz\)

Hệ số: 5

Phần biến: \(xyz\)

Bậc: 1+1+1=3

- \(-xyz\cdot\dfrac{2}{3}y=-\dfrac{2}{3}xy^2z\)

Hệ số: \(-\dfrac{2}{3}\)

Phần biến: \(xy^2z\)

Bậc: 1+2+1=4

- \(-2x^2\left(-\dfrac{1}{6}\right)x=\dfrac{1}{3}x^3\)

Hệ số: \(\dfrac{1}{3}\)

Biến: \(x^3\)

Bậc: 3

\(=\frac{6xyz^2+9xyz^2-4xyz^2}{12}=\frac{3xyz^2}{4}\)

a)tổng :\(=-\dfrac{1}{3}x^3y^4+\dfrac{4}{3}x^3y^4=\dfrac{3}{3}x^3y^4=x^3y^4\)

hiệu :\(=-\dfrac{1}{3}x^3y^4-\dfrac{4}{3}x^3y^4=-\dfrac{5}{3}x^3y^4\)

b)tôngr : \(=\left(-2+1\right)xyz=-xyz\)

hiệu :\(=\left(-2-1\right)xyz=-3xyz\)

`A = x - 2y + xy - 3x + y^2`

Bậc: `2`.

`B = (1-1/2)xyz - x^2y + (1+1/2)xz`

`= 1/2xyz - x^2y + 3/2xz`

Bậc: `3`

a) \(2x=5y\)⇒\(x=\dfrac{5}{2}y\)⇒\(xy=\dfrac{5}{2}y^2\)

Thay \(xy=250\), ta có:

\(250=\dfrac{5}{2}y^2\)

⇒\(y^2=100\)⇒\(y=+-10\)

+) \(y=10\text{⇒}x=250:10=25\)

+) \(y=-10\text{⇒}x=250:-10=-25\)

\(a,2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=k\\ \Rightarrow x=5k;y=2k\\ xy=250\Rightarrow5k\cdot2k=250\Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=25;y=10\\x=-25;y=-10\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{4}=a\Rightarrow x=3a;y=2a;z=4a\\ xyz=192\Rightarrow24a^3=192\Rightarrow a^3=8\Rightarrow a=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=4\\z=8\end{matrix}\right.\\ c,\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=q\Rightarrow x=5q;y=2q;z=-3q\\ xyz=240\Rightarrow-30q^3=240\Rightarrow q^3=-8\Rightarrow q=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-10\\y=-4\\z=6\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}2x=5y\\xy=250\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5y=0\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2xy-5y^2=0\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y^2=500\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10\\x=25\end{matrix}\right.\)

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\)

=>\(\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}=k\)

=>x+1=4k; y-2=2k; z+2=3k

=>x=4k-1; y=2k+2; z=3k-2

xyz=12

=>(4k-1)(2k+2)(3k-2)=12

=>(4k-1)(k+1)(3k-2)=6

=>(4k-1)(3k^2-2k+3k-2)=6

=>(3k^2+k-2)(4k-1)=6

=>12k^3-3k^2+4k^2-k-8k+2-6=0

=>12k^3+k^2-9k-7=0

=>

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\)

=>\(\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}=k\)

=>x+1=4k; y-2=2k; z+2=3k

=>x=4k-1; y=2k+2; z=3k-2

xyz=12

=>(4k-1)(2k+2)(3k-2)=12

=>(4k-1)(k+1)(3k-2)=6

=>(4k-1)(3k^2-2k+3k-2)=6

=>(3k^2+k-2)(4k-1)=6

=>12k^3-3k^2+4k^2-k-8k+2-6=0

=>12k^3+k^2-9k-4=0

=>k=1

=>x=4k-1=3; y=2k+2=4; z=3k-2=3-2=1

Phân thức số 2 có thật sự là $\frac{z}{y-2}$ không bạn? Bạn xem lại đề.