a, ĐKXĐ:\(x\ne0,x\ne2\)

\(\dfrac{2}{x-2}-\dfrac{1}{x}=\dfrac{3}{x\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{3}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x-x+2-3}{x\left(x-2\right)}=0\\ \Rightarrow x-1=0\\ \Leftrightarrow x=1\left(tm\right)\)

b, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{1}{x+3}-\dfrac{2x-1}{x-3}=\dfrac{x^2-15}{x^2-9}\\ \Leftrightarrow\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x-1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-15}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{x-3-\left(2x^2-x+6x-3\right)-\left(x^2-15\right)}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow x-3-2x^2+x-6x+3-x^2+15=0\\ \Leftrightarrow-3x^2-4x+15=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)

\(x^4-2x^3+4x^2-3x+2=0\\ \Leftrightarrow x^4-2x^3+x^2+3x^2-3x+2=0\\ \Leftrightarrow x^2\left(x^2-2x+1\right)+\left(3x^2-3x+2\right)=0\\ \Leftrightarrow x^2\left(x-1\right)^2+\left(3x^2-3x+2\right)=0\)

Vì \(x^2\left(x-1\right)^2\ge0\) và dễ dàng chứng minh được \(3x^2-3x+2>0\) nên pt vô nghiệm

Diện tích xung quanh hình lập phương là:

\(4,2\times4,2\times4=70,56\left(m^2\right)\)

Diện tích toàn phần hình lập phương là:

\(4,2\times4,2\times6=105,84\left(m^2\right)\)

Thể tích hình lập phương là:

\(4,2\times4,2\times4,2=74,088\left(m^3\right)\)

\(\dfrac{4}{5}+\dfrac{1}{3}=\dfrac{12}{15}+\dfrac{5}{15}=\dfrac{17}{15}\\ \dfrac{13}{7}-\dfrac{7}{7}=\dfrac{6}{7}\\ \dfrac{4}{9}\times\dfrac{2}{5}=\dfrac{8}{45}\\ \dfrac{3}{2}:4=\dfrac{3}{2}\times\dfrac{1}{4}=\dfrac{3}{8}\)

\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)

\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 7,x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

\(8,x^2\left(x-2y\right)+3x\left(x-2y\right)=\left(x^2+3x\right)\left(x-2y\right)=x\left(x+3\right)\left(x-2y\right)\\ 9,\left(5x+2\right)\left(x-3\right)-x\left(x-3\right)=\left(x-3\right)\left(5x+2-x\right)=\left(x-3\right)\left(4x+2\right)=2\left(x-3\right)\left(2x+1\right)\\ 10,\left(5x-3\right)\left(x+2\right)-2x\left(x+2\right)=\left(x+2\right)\left(5x-3-2x\right)=\left(x+2\right)\left(3x-3\right)=3\left(x+2\right)\left(x-1\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)

1, Hoành độ giao điểm 2 đường thẳng đó là:

\(2x-3=x+1\Leftrightarrow x=4\)

Tung độ giao điểm 2 đường thẳng đó là:

\(y=2x-3=2.1-3=-1\)

Vậy tọa độ giao điểm 2 đường thẳng đó là:\(\left(4;-1\right)\)

2, Để đường thẳng (d₁) đi qua A(1;-2) thì:

\(-2=\left(2m-1\right).1+n+2\\ \Leftrightarrow2m-1+n+2+2=0\\ \Leftrightarrow2m+n+3=0\left(1\right)\)

Để đường thẳng (d₂) đi qua A(1;-2) thì:

\(-2=2n.1+2m-3\\ \Leftrightarrow2n+2m-3+2=0\\ \Leftrightarrow2n+2m-1=0\left(2\right)\)

Từ (1), (2) ta có hệ: \(\left\{{}\begin{matrix}2m+n+3=0\\2n+2m-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-\dfrac{7}{2}\\n=4\end{matrix}\right.\)

\(\dfrac{2x-3}{8}=\dfrac{x+6}{12}\\ \Leftrightarrow\dfrac{3\left(2x-3\right)}{24}-\dfrac{2\left(x+6\right)}{24}=0\\ \Leftrightarrow6x-9-2x-12=0\\ \Leftrightarrow4x-21=0\\ \Leftrightarrow x=\dfrac{21}{4}\)

\(\left|-3x-5\right|-x-3=0\\ \Leftrightarrow\left|-3x-5\right|=x+3\left(x\ge-3\right)\\ \Leftrightarrow\left[{}\begin{matrix}-3x-5=x+3\\-3x-5=-x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

a, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a,\dfrac{1}{y}=b\)

Hệ \(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\left(tm\right)\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}-\dfrac{2y}{2}=\dfrac{2}{2}\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1-2y=2\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2y=3\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)