Bài 10: Chia đơn thức cho đơn thức

a, \(\dfrac{3a^2b-4ab^2}{5ab}=\dfrac{ab\left(3a-4b\right)}{5ab}=\dfrac{3a-4b}{5}\)

b, \(\dfrac{3x^3y^2-5x^2y^3+4x^3y^3}{x^2y^2}=\dfrac{x^2y^2\left(3x-5y+4xy\right)}{x^2y^2}\)

\(=3x-5y+4xy\)

c, \(\dfrac{2a^5b^4+3a^4b^3}{-3a^4b^5}=\dfrac{a^4b^3\left(2ab+3\right)}{-3a^4b^5}=\dfrac{2ab+3}{-3b^2}\)

d, \(\dfrac{-a^5b^4+3a^6b^2}{4a^4b^2}=\dfrac{-a^4b^2\left(ab^2+3a^2\right)}{4a^4b^2}=\dfrac{-\left(ab^2+3a^2\right)}{4}\)

Chúc bạn học tốt!!!

a. \(\left(3a^2b-4ab^3\right):5ab=3a^2b:5ab-4ab^3:5ab=\dfrac{3}{5}a-\dfrac{4}{5}b^2\)

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

c. \(\left(2a^5b^4+3a^4b^3\right):\left(-3a^4b^5\right)=2a^5b^4:\left(-3a^4b^5\right)+3a^4b^3:\left(-3a^4b^5\right)=-\dfrac{2a}{3b}-\dfrac{1}{b^2}\)

d. \(\left(-a^5b^4+3a^6b^2\right):4a^4b^2=\left(-a^5b^4\right):4a^4b^2+3a^6b^2:4a^4b^2=-\dfrac{1ab^2}{4}+\dfrac{3a^2}{4}\)

$5(x-y)^5:z(x-y)^2$

$=(5:z).[(x-y)^5:(x-y)^2]$

$=\dfrac{5}{z}.(x-y)^3$

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 \(\left\{{}\begin{matrix}n-2>0\\n-4>0\end{matrix}\right.\Leftrightarrow n>4\)

giải

(-12)^3:8^3

=(-12:8)^3

=(-1,5)^3

chúc bạn học tốt nha

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

Chúc bạn học tốt!!!

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

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 :))

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

Bài 3: 

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

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

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

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

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

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