ĐK : \(a\ne b\ne c\)

\(\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

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

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca\right)-3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{2\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)

\(=\dfrac{a+b+c}{2}\)

a) \(2x^2y^3.\dfrac{1}{4}xy^3\left(-3\right)xy\)

\(=\left(-3.2.\dfrac{1}{4}\right)x^4y^7\)

\(=\dfrac{-3}{2}x^4y^7\)

\(\Rightarrow Hệ\) số: \(\dfrac{-3}{2}\)

Phần biến: \(x^4y^7\)

b) \(\left(-2x^3y\right)^2.xy^2.\dfrac{1}{5}y^5\)

\(=\dfrac{4}{5}x^7y^9\)

\(\Rightarrow Phần\) biến: \(x^7y^9\)

Hệ số: \(\dfrac{4}{5}.\)

a/ \(2x^2y^3\cdot\dfrac{1}{4}xy^3\left(-3xy\right)\)

\(=\left[2\cdot\dfrac{1}{4}\cdot\left(-3\right)\right]\left(x^2.x.x\right)\left(y^3.y^3.y\right)\)

\(=-\dfrac{3}{2}x^4y^7\)

Phần biến: \(x^4y^7\)

Hệ số: \(-\dfrac{3}{2}\)

b/ \(\left(-2x^3y\right)^2\cdot xy^2\cdot\dfrac{1}{5}y^5=4x^6y^2\cdot xy^2\cdot\dfrac{1}{5}y^5\) \(=4\cdot\dfrac{1}{5}\left(x^6\cdot x\right)\left(y^2\cdot y^2\cdot y^5\right)=\dfrac{4}{5}x^7y^9\)

Phần biến: \(\dfrac{4}{5}\)

Hệ số: \(x^7y^9\)

XEM CÓ SAI ĐỀ BÀI KHÔNG, MK RÚT GỌN RA TO LẮM

\(=\dfrac{x+5\sqrt{x}+6-x+5\sqrt{x}-6}{\left(\sqrt{x}+3\right)^2\cdot\left(\sqrt{x}-3\right)}\cdot\dfrac{x-9}{\sqrt{x}}\)

\(=\dfrac{10\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{10}{\sqrt{x}+3}\)

a/\(\frac{10x}{5x^2}=\frac{2}{x}\)

b/\(\frac{x\left(x^2-y^2\right)}{x^2\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}=\frac{x-y}{x}\)

1) \(\dfrac{x^2-18x-19}{x^2-1}=\dfrac{x^2-19x+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{x\left(x-19\right)+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-19\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-19}{x-1}\)

2) \(\dfrac{x\left(4x^2-8x+4\right)}{2x^3-2x^2}=\dfrac{4x\left(x^2-2x+1\right)}{2x^2\left(x-1\right)}=\dfrac{4x\left(x-1\right)^2}{2x^2\left(x-1\right)}=\dfrac{2\left(x-1\right)}{x}\)

1.=\(\dfrac{(x^2+x)-(19x+19)}{(x+1)(x-1)}\)

=\(\dfrac{x(x+1)-19(x+1)}{(x+1)(x-1)}\)

=\(\dfrac{(x+1)(x-19)}{(x+1)(x-1)}\)

=\(\dfrac{x-19}{x-1}\)

\(=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{x+2-x^2}{x+2}-\dfrac{x^2+6x+4}{x}\)

\(=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)-x^2-6x-4}{x}\)

\(=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-6x-4}{x}\)

\(=\dfrac{-x^3-2x^2-2x}{x}=-x^2-2x-2\)

ĐKXĐ : \(x^2-5x\ne0\Leftrightarrow x\left(x-5\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne5\end{cases}}\)

a) \(A=\frac{x^2-10x+25}{x^2-5x}\)

\(A=\frac{\left(x-5\right)^2}{x\left(x-5\right)}\)

\(A=\frac{x-5}{x}\)

b) Để phân thức bằng 0 thì \(x-5=0\Leftrightarrow x=5\)

Mà ĐKXĐ \(x\ne5\)=> ko có giá trị của x để phân thức bằng 0

c) Để phân thức bằng 0 thì :

\(\frac{x-5}{x}=\frac{5}{2}\)

\(2x-10=5x\)

\(-10=3x\)

\(x=\frac{-3}{10}\)

a,\(\frac{x^2-10x+25}{x^2-5x}=\frac{\left(x-5\right)^2}{x\left(x-5\right)}=\frac{x-5}{x}\)

b,Để phân thức có giá trị bằng 0 thì \(\frac{x-5}{x}=0\)

Mà: Theo điều kiện ta có: \(x\ne0\)

nên để: \(\frac{x-5}{x}=0\)thì: \(x-5=0\Leftrightarrow x=5\)

c,Để phân thức có giá trị bằng 5/2 thì:

\(\frac{x-5}{x}=\frac{5}{2}\)

\(\Leftrightarrow2\left(x-5\right)=5x\)

\(\Leftrightarrow2x-10=5x\)

\(\Leftrightarrow2x-5x=10\)

\(\Leftrightarrow-3x=10\Rightarrow x=-\frac{10}{3}\)

=.= hk tốt!!