Bài 7: Phép nhân các phân thức đại số

Gọi phân thức cần tìm là A

Theo đề, ta có: \(\dfrac{x^3-x^2}{1-x^2}=\dfrac{A}{x+1}\)

=>\(\dfrac{-x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{A}{x+1}\)

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

=>\(A=-x^2\)

a: \(A=\dfrac{x}{x+5}-\dfrac{7x-15}{25-x^2}+\dfrac{3}{x-5}\)

\(=\dfrac{x}{x+5}+\dfrac{7x-15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3}{x-5}\)

\(=\dfrac{x\left(x-5\right)+7x-15+3\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{x^2-5x+7x-15+3x+15}{\left(x+5\right)\left(x-5\right)}=\dfrac{x^2+5x}{\left(x+5\right)\left(x-5\right)}=\dfrac{x}{x-5}\)

b: |2x-1|=9

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(loại\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Khi x=-4 thì \(A=\dfrac{-4}{-4-5}=\dfrac{-4}{-9}=\dfrac{4}{9}\)

c: \(P=A\cdot B=\dfrac{x}{x-5}\cdot\left(x^2-2x-15\right)\)

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

\(=x^2+3x\)

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

Dấu = xảy ra khi x=-3/2

a) ĐKXĐ:

x³ - 1 khác 0

x khác 1

b) A = (5x² + 5x + 5)/(x³ - 1)

= 5(x² + x + 1)/[(x - 1)(x² + x + 1)]

= 5/(x - 1)

Thay x = 7 vào A, ta được:

A = 5/(7 - 1)

= 5/6

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)

a: A=(3x+5)(2x-7)

=6x^2-21x+10x-35

=6x^2-11x-35

b: (x^2-xy+y^2)(x+y)=x^3+y^3

a: A=(x-7)(x-5)=x^2-5x-7x+35=x^2-12x+35

b: A=(x^2-2x+1)(x-1)

=(x-1)^2*(x-1)

=(x-1)^3

=x^3-3x^2+3x-1

a: 4(x-1)(x+1)

=4(x^2-1)

=4x^2-4

b: (x+3)(x^2+3x-5)

=x^3+3x^2-5x+3x^2+9x-15

=x^3+6x^2+4x-15

a: (xy-1)(xy+5)

=x^2y^2+5xy-xy-5

=x^2y^2+4xy-5

b: (x^2-2x+3)(x-5)

=x^3-5x^2-2x^2+10x+3x-15

=x^3-7x^2+13x-15

Giúp mik vs mik đang cần gấp:(())

a: \(=\dfrac{24y^5}{-12y^3}\cdot\dfrac{21x}{7x^2}=-\dfrac{6y^2}{x}\)

b: \(=\dfrac{2x}{3}\cdot\dfrac{6x^2}{5}=\dfrac{4x^3}{5}\)

\(\dfrac{3x-1}{10x^2+2x}\cdot\dfrac{25x^2+10x+1}{1-9x^2}\)

\(=\dfrac{3x-1}{2x\left(5x+1\right)}\cdot\dfrac{\left(5x+1\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

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

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

\(=\dfrac{-\left(5x+1\right)}{2x\left(1+3x\right)}\)

\(=\dfrac{-5x-1}{2x+6x^2}\)

`1)[x^3-8]/[5x+20].[x^2+4x]/[x^2+2x+4]`

`=[(x-2)(x^2+2x+4)]/[5(x+4)].[x(x+4)]/[x^2+2x+4]`

`=[x(x-2)]/5=[x^2-2x]/5`

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`2)[x+1]/[x^2-2x-8].[4-x]/[x^2+x]`

`=[x+1]/[(x-4)(x+2)].[-(x-4)]/[x(x+1)]`

`=[-1]/[x(x+2)]=[-1]/[x^2+2x]`

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`3)[3x^2-x]/[x^2-1].[1-x^4]/[(1-3x)^3]`

`=[x(3x-1)]/[x^2-1].[-(x^2-1)(x^2+1)]/[-(3x-1)(9x^2+3x+1)]`

`=[x(x^2+1)]/[9x^2+3x+1]=[x^3+x]/[9x^2+3x+1]`

a) \(3x\left(5x^2-2x-1\right)\)

\(=3x.5x^2-3x.2x+3x.\left(-1\right)\)

\(=15x^3-6x^2-3x\)

b) \(\left(x^3-2xy+3\right)\left(-xy\right)\)

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

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

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

mấy câu sau vt lại đè

=15x²-9xy