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\(=\left(\dfrac{y}{x\left(y-2x\right)}-\dfrac{2}{y\left(y+1\right)-2x\left(y+1\right)}\right)\cdot\left(1+y\right)\)
\(=\left(\dfrac{y}{x\left(y-2x\right)}-\dfrac{2}{\left(y+1\right)\left(y-2x\right)}\right)\cdot\left(y+1\right)\)
\(=\left(\dfrac{y\left(y+1\right)-2x}{x\left(y-2x\right)\left(y+1\right)}\right)\cdot\dfrac{y+1}{1}\)
\(=\dfrac{y^2+2y-2x}{x\left(y-2x\right)}\)
Với điều kiện xy\(\ne\)0;+ -3/2 y;x\(\ne\)-y các phân thức có nghĩa. Ta có
\(\frac{5x\left(2x-3y\right)^2}{3y\left(4x^2-9y^2\right)}:\frac{\left(2x^2+2xy\right)\left(2x-3y\right)}{2x^2y+5xy^2+3y^3}\)\(=\)\(\frac{5x\left(2x-3y\right)^2.y\left(2x^2+5xy+3y^2\right)}{3y\left(4x^2-9y^2\right).2x\left(x+y\right).\left(2x-3y\right)}\)
\(=\)\(\frac{10xy\left(2x-3y\right)^2.\left(2x^2+2xy+3xy+3y^2\right)}{6xy\left(2x-3y\right).\left(2x+3y\right)\left(x+y\right)\left(2x-3y\right)}\)\(=\)\(\frac{10xy\left(2x-3y\right)^2\left(x+y\right).\left(2x+3y\right)}{6xy\left(2x-3y\right)^2.\left(2x+3y\right).\left(x+y\right)}\)
\(=\)\(\frac{5}{3}\)
ĐK \(\hept{\begin{cases}xy\ne0\\2x-3y\ne0,2x+3y\ne0\\x\ne-y\end{cases}}\)
\(=\frac{5x\left(2x-3y\right)^2}{3y\left(2x+3y\right)\left(2x-3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{xy\left(2x+3y\right)+y^2\left(2x+3y\right)}\)
\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{\left(2x+3y\right)\left(xy+y^2\right)}\)
\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}.\frac{y\left(x+y\right)\left(2x+3y\right)}{2x\left(x+y\right)\left(2x-3y\right)}=\frac{5}{6}\)
Vậy giá trị của biểu thức không phụ thuộc vào biến
\(a.\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{-2\left(y-x\right)}{3\left(y-x\right)}=\dfrac{-2}{3}\)
\(b.\dfrac{x-2}{-x}=\dfrac{2-x}{x}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{8-x^3}{x\left(x^2+2x+4\right)}\)
\(\dfrac{3x}{x+y}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{-3x\left(x-y\right)}{\left(x+y\right)\left(y-x\right)}=\dfrac{-3x\left(x-y\right)}{y^2-x^2}\)
c: \(\dfrac{-3x\left(x-y\right)}{y^2-x^2}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{3x}{x+y}\)
a: \(\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{2\left(x-y\right)}{-3\left(x-y\right)}=\dfrac{-2}{3}\)
b: \(\dfrac{8-x^3}{x\left(x^2+2x+4\right)}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{2-x}{x}\)
a)2(x-y)/(-3)(x-y)=-2/3
b)8-x^3=(2-x)(x^2+2x+4) => Vế phải =(2-x)/x=(x-2)/-x
c)y^2-x^2=(y+x)(y-x) bạn đổi dấu rồi rút gọn là được,cũng tương tự như trên ý
1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)
\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{2x-2xy-3+3y}{1-3y+3y^2-y^3}=\dfrac{2x\left(1-y\right)-3\left(1-y\right)}{\left(1-y\right)^3}\)
\(=\dfrac{\left(2x-3\right)\left(1-y\right)}{\left(1-y\right)^3}=\dfrac{2x-3}{\left(1-y\right)^2}\)