Ta có:

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

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

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

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

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

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

\(=\dfrac{\left(2x-2y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{2x-2y}{x^2+xy+y^2}\)

Bạn viết đề cẩn thận bằng công thức toán thì sẽ tăng khả năng nhận được sự giúp đỡ hơn. Viết như thế này nhìn rối mắt cực. 

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

b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)

\(=\dfrac{\left(x+y\right)^2}{x^2+xy+x^2-y^2}=\dfrac{\left(x+y\right)^2}{x\left(x+y\right)+\left(x-y\right)\left(x+y\right)}\)

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

c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)

\(=ax\left(x-a\right)\)

Giải:
Ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

+) \(\frac{x}{10}=-7\Rightarrow x=-70\)

+) \(\frac{y}{15}=-7\Rightarrow y=-105\)

+) \(\frac{z}{12}=-7\Rightarrow z=-84\)

Vậy...

Cách 1:

\(\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{4}\Leftrightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)

Từ (1) và (2) suy ra:\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng Tc dãy tỉ số bằng nhau ta có:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

\(\Rightarrow\begin{cases}\frac{x}{10}=-7\Rightarrow x=-70\\\frac{y}{15}=-7\Rightarrow y=-105\\\frac{z}{12}=-7\Rightarrow z=-84\end{cases}\)

MK biết làm bằng 1 cách à bn

Bạn tham khảo xem sao

x/2=y/3;y/5=z/4

→x/10=y/15;y/15=z/12

 →x/10=y/15=z/12

Áp dụng t/x dãy tỉ số bằng nhau

ta có:x/10=y/15=z/12=x-y+z/10-15+12=-49/7=-7

Suy ra:x=-7.10=-70          y=-7.15=-105      z=-7.12=-84

\(x^3+3x^2y+3xy^2+y^3-x-y=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)

Ta có : \(x^3+3x^2y+3xy^2+y^3-x-y.\)

\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)

a, (\(x\) + y).(\(x\) + y)² - 3\(xy\).(\(x\) + y) 

= (\(x+y\))³ - 3\(x^2\)y - 3\(xy^2\)

= \(x^3\) + 3\(x^2\).y + 3\(xy^2\) + y³ - 3\(x^2\).y  - 3\(xy^2\)

= \(x^3\) + y³ 

b, (\(x-y\)).(\(x-y\))² - 3\(xy\).(\(x-y\)) 

=    (\(x\) - y)³ - 3\(x^2\).y + 3\(xy^2\)

= \(x^3\) - 3\(x^2\)y + 3\(xy^2\) - y³ - 3\(x^2\)y + 3\(xy^2\)

= \(x^3\) - 6\(x^2\)y + 6\(xy^2\) - y³

c, (\(x\) - 2y)² + 4y²

=  \(x^2\) - 4\(xy\) + 4y² + 4y²

=  \(x^2\) - 4\(xy\) + 8y²

\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

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

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)

a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)

b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)

\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)

\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)

\(=\left(5y-3x\right)\left(5y+3x\right)\)

\(=25y^2-9x^2\)

d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)

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

\(=\dfrac{9}{4}y^2-x^2\)

e: \(\left(a+b+c\right)\left(a+b-c\right)\)

\(=\left(a+b\right)^2-c^2\)

\(=a^2+2ab+b^2-c^2\)

\(A=x^3+3xy\left(x+y\right)+y^3-xy\left(x+y\right)+x^2+y^2+xy+2\)

\(A=\left(x+y\right)^3-xy.\left(-1\right)+x^2+y^2+xy+2\)

\(A=\left(-1\right)^3+x^2+y^2+2xy+2\)

\(A=\left(x+y\right)^2+1\)

\(A=\left(-1\right)^2+1=2\)