Bài 10: Chia đơn thức cho đơn thức

Bài 3: 

\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)

\(=\left(x^2-9\right)\left(x^2-1\right)+15\)

\(=x^4-10x^2+9+15\)

\(=x^4-10x^2+24\)

\(=\left(x^2-4\right)\left(x^2-6\right)\)

\(=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)

Do \(y(y+x)\ne0 \) nên \(y\ne0;y\ne-x\)

Đặt \(t=\dfrac{x}{y},t\ne-1\)

Ta có: \(x^2-xy=2y^2 \Rightarrow(\dfrac{x}{y})^2-\dfrac{x}{y}=2\)

\(\Rightarrow t^2-t-2=0 \Leftrightarrow t=2 \ \ vì \ \ t\ne-1\)

\(\Rightarrow A=\dfrac{1007\dfrac{x}{y}-1}{\dfrac{x}{y}+2012}=\dfrac{2013}{2014}\)

cách khác

\(\left\{{}\begin{matrix}x^2-xy=2y^2\left(1\right)\\y\left(x+y\right)\ne0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(x^2-y^2\right)-\left(xy+y^2\right)=\left(x-y\right)\left(x+y\right)-y\left(x+y\right)=\left(x+y\right)\left(x-2y\right)=0\)

Từ (2) =>\(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)

\(A=\dfrac{1007x-y}{x+2012y}=\dfrac{1007.2y-y}{2y+2012y}=\dfrac{\left(1007.2-1\right)y}{\left(2+2013\right)y}=\dfrac{2013y}{2014y}\)

Từ (2)=> \(y\ne0\) \(\Rightarrow A=\dfrac{2013}{2014}\)

làm giùm mình đi mình đang cần gấp

1) GIẢI

\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\sqrt{3}\)

\(\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=\left(\sqrt{3}\right)^2=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{zx}=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{x+y+z}{xyz}\right)=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{xyz}{xyz}\right)=3\) ( Do x+y+z = xyz )

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.1=3\)

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

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\)

Chúc bạn học tốt :))

a,

\(\dfrac{18\left(x-y\right)^{10}}{2\left(x-y\right)^5}=9\left(x-y\right)^5\)

b, \(\dfrac{10\left(x-2\right)^{12}}{\left(2-x\right)^{10}}=\dfrac{10\left(x-2\right)^{12}}{\left(x-2\right)^{10}}=10\left(x-2\right)^2\)

c, \(\dfrac{-18\left(x-3\right)^5}{2\left(3-x\right)^3}=\dfrac{-18\left(x-3\right)^5}{-2\left(x-3\right)^3}=9\left(x-3\right)^2\)

d,\(\dfrac{x^2-6x+9}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\)

e, \(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-2x+x-2}{x+1}=\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=x-2\)

\(\dfrac{\left(-xyz^2\right)^3}{\left(-x^2yz^3\right)^2}=\dfrac{x^3y^3z^6}{x^4y^2z^6}=\dfrac{y}{x}\)

Chúc bạn học tốt!!!

giải

(-12)^3:8^3

=(-12:8)^3

=(-1,5)^3

chúc bạn học tốt nha

a: \(5x^ny^3:4x^2y^2=\dfrac{5}{4}x^{n-2}y\)

Để đây là phép chia hết thì n-2>0

hay n>2

b: \(x^ny^{n+1}:x^2y^5=x^{n-2}y^{n-4}\)

Để đây là phép chia hết thì \(\left\{{}\begin{matrix}n-2>0\\n-4>0\end{matrix}\right.\Leftrightarrow n>4\)

\(R=x^2+y^2-x+6y+10\)

\(=\left(x^2-x+10\right)+\left(y^2+6y\right)\)

\(=\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+10\right)+\left(y^2+2.y.3+3^2-9\right)\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{39}{4}+\left(y+3\right)^2-9\)

\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\left(y+3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)

\(\Rightarrow R\ge\dfrac{3}{4}\forall x,y\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

Vậy ...

Ta có:

\(R=x^2+y^2-x+6y+10\)

\(R=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}+y^2+3y+3y+9+\dfrac{3}{4}\)

\(R=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Hay \(R\ge\dfrac{3}{4}\) với mọi giá trị của \(x;y\in R\).

Để \(R=\dfrac{3}{4}\) thì \(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\Vậy..............

Chúc bạn học tốt!!!

Tương tự: Câu hỏi của Nguyễn Thị Kim Anh - Toán lớp 8 | Học trực tuyến

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