Bài 4: Quy đồng mẫu thức nhiều phân thức

\(=x^2+4x-21-x^2-4x+5=-16\)

Lần sau bạn lưu ý viết đầy đủ yêu cầu đề.

Coi đây là bài toán rút gọn.

Lời giải:
$(x-3)(x+7)-(x+5)(x-1)$
$=(x^2+7x-3x-21)-(x^2-x+5x-5)$

$=(x^2+4x-21)-(x^2+4x-5)$

$=-21+5=-16$

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

h: \(\dfrac{a}{a-b}=\dfrac{a}{a-b}\)

\(\dfrac{b}{b-a}=\dfrac{-b}{a-b}\)

\(a,PT\left(1\right)=\dfrac{75y^4}{42x^2y^5};PT\left(2\right)=\dfrac{28x}{42x^2y^5}\\ b,PT\left(1\right)=\dfrac{11y^2}{102x^4y^3};PT\left(2\right)=\dfrac{9x^3}{10x^4y^3}\\ c,PT\left(1\right)=\dfrac{3x\left(3x+1\right)}{36x^2y^4};PT\left(2\right)=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ d,PT\left(1\right)=\dfrac{6y^2}{36x^3y^4};PT\left(2\right)=\dfrac{4x\left(x+1\right)}{36x^3y^4};PT\left(3\right)=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\)

\(e,PT\left(1\right)=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};PT\left(2\right)=\dfrac{75x^2y^3}{120x^4y^5};PT\left(3\right)=\dfrac{8x^3}{120x^4y^5}\\ f,PT\left(1\right)=\dfrac{3\left(x+1\right)\left(4x-4\right)}{6x\left(x+3\right)\left(x+1\right)};PT\left(2\right)=\dfrac{2\left(x+3\right)\left(x-3\right)}{6x\left(x+1\right)\left(x+3\right)}\)

\(g,PT\left(1\right)=\dfrac{4x^2}{2x\left(x+2\right)^3};PT\left(2\right)=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\\ h,PT\left(1\right)=\dfrac{5}{3x\left(x-2\right)\left(x+2\right)}=\dfrac{10\left(x+3\right)}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\\ PT\left(2\right)=\dfrac{3}{2\left(x+2\right)\left(x+3\right)}=\dfrac{9x\left(x-2\right)}{6x\left(x+2\right)\left(x+3\right)\left(x-2\right)}\)

\(x^2+1=\dfrac{x^4-1}{x^2-1}\)

Lời giải:
$\frac{4x^2-3x+8}{x^3-1}$
$\frac{2x}{x^2+x+1}=\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{2x^2-2x}{x^3-1}$

$\frac{6}{1-x}=\frac{-6(x^2+x+1)}{(x-1)(x^2+x+1)}=\frac{-6x^2-6x-6}{x^3-1}$