Có \(\frac{x}{2}=\frac{y}{4}\Rightarrow\frac{x}{y}=\frac{2}{4}\Rightarrow x=2;y=4\)

Thế x;y vào biểu thức P ta có :

\(P=\frac{7x^2+3y^2}{14x^2-3y^2}=\frac{7\cdot2^2+3\cdot4^2}{14\cdot2^2-3\cdot4^2}=\frac{19}{2}\)

Vậy \(P=\frac{19}{2}\)

Bài 2:

Giải:

Ta có: \(\frac{x}{2}=\frac{y}{4}\Rightarrow\frac{x}{1}=\frac{y}{2}\)

Đặt \(\frac{x}{1}=\frac{y}{2}=k\Rightarrow\left\{\begin{matrix}x=k\\y=2k\end{matrix}\right.\)

Lại có: \(P=\frac{7x^2+3y^2}{14x^2-3y^2}=\frac{7k^2+12k^2}{14k^2-12k^2}=\frac{19}{2}=9,5\)

Vậy P = 9,5

Với điều kiện xy\(\ne\)0;+ -3/2 y;x\(\ne\)-y các phân thức có nghĩa. Ta có

\(\frac{5x\left(2x-3y\right)^2}{3y\left(4x^2-9y^2\right)}:\frac{\left(2x^2+2xy\right)\left(2x-3y\right)}{2x^2y+5xy^2+3y^3}\)\(=\)\(\frac{5x\left(2x-3y\right)^2.y\left(2x^2+5xy+3y^2\right)}{3y\left(4x^2-9y^2\right).2x\left(x+y\right).\left(2x-3y\right)}\)

\(=\)\(\frac{10xy\left(2x-3y\right)^2.\left(2x^2+2xy+3xy+3y^2\right)}{6xy\left(2x-3y\right).\left(2x+3y\right)\left(x+y\right)\left(2x-3y\right)}\)\(=\)\(\frac{10xy\left(2x-3y\right)^2\left(x+y\right).\left(2x+3y\right)}{6xy\left(2x-3y\right)^2.\left(2x+3y\right).\left(x+y\right)}\)

\(=\)\(\frac{5}{3}\)

ĐK \(\hept{\begin{cases}xy\ne0\\2x-3y\ne0,2x+3y\ne0\\x\ne-y\end{cases}}\)

\(=\frac{5x\left(2x-3y\right)^2}{3y\left(2x+3y\right)\left(2x-3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{xy\left(2x+3y\right)+y^2\left(2x+3y\right)}\)

\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{\left(2x+3y\right)\left(xy+y^2\right)}\)

\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}.\frac{y\left(x+y\right)\left(2x+3y\right)}{2x\left(x+y\right)\left(2x-3y\right)}=\frac{5}{6}\)

Vậy giá trị của biểu thức không phụ thuộc vào biến

\(=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)

Bài 1:

a) \(2x\left(x^2-5x+6\right)=2x^3-10x^2+12x\)

b) \(\left(7x^5+14x^2y^3-28x^3y^2\right):7x^2=x^3+2y^3-4xy^2\)

Bài 2:

\(x^2+y^2+2x-8y+17=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

\(a,=7xy\left(x^2-2xy+y^2\right)=7xy\left(x-y\right)^2\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\)

`a)7x^3y^2+14x^2y^3+7xy^4`

`=7xy^2(x^2+2xy+y^2)`

`=7xy^2(x+y)^2`

______________________________________________

`b)x^2-xy+5x-5y`

`=x(x-y)+5(x-y)`

`=(x-y)(x+5)`

______________________________________________

`c)3x^2-6xy-12+3y^2`

`=3(x^2-2xy-4+y^2)`

`=3[(x-y)^2-4]`

`=3(x-y-2)(x-y+2)`

a)7x³y²+14x²y³+7xy⁴

=7xy²(x²+2xy+y²)

=7xy²(x+y)²

b)x²-xy + 5x - 5y

=x(x-y) + 5(x-y)

=(x-y) (x+5)

p

Em kiểm tra lại đề bài nhé vì:

\(Q=\left(x^3.x.y^n.y-\frac{1}{2}x^3.y^n.y^2\right):\frac{1}{2}x^3y^n-\left(4.5.x^2.x^2.y\right):\left(5x^2y\right)\)

\(=x^3y^n\left(xy-\frac{1}{2}y^2\right):\frac{1}{2}x^3y^n-5x^2y\left(4x^2\right):5x^2y\)

\(=2xy-y^2-4x^2=-\left(x^2-2xy+y^2\right)-3x^2=-\left[\left(x-y\right)^2+3x^2\right]< 0\)Với mọi x, y khác 0

=> Q luôn có gia trị âm với mọi x, y khác 0.