Bài 6: Phép trừ các phân thức đại số

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

\(=\dfrac{1}{x\left(1-5x\right)}+\dfrac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)

\(=\dfrac{1+5x+25x^2-15x}{x\left(1-5x\right)\left(1+5x\right)}\)

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

\(\dfrac{x+1}{x+4}-\dfrac{x^2-4}{x^2-16}\\ =\dfrac{x+1}{x+4}-\dfrac{x^2-4}{\left(x-4\right)\left(x+4\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)-x^2+4}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{x^2+x-4x-4-x^2+4}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{-3x}{\left(x+4\right)\left(x-4\right)}\)

a: Sửa đề: \(\dfrac{1}{2x}+\dfrac{x-2}{3x-3}+\dfrac{x}{x^2-x}\)

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

\(=\dfrac{3\left(x-1\right)+2x^2-4x+6x}{6x\left(x-1\right)}\)

\(=\dfrac{3x-3+2x^2+2x}{6x\left(x-1\right)}=\dfrac{2x^2+5x-3}{6x\left(x-1\right)}\)

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

\(=\dfrac{2x-6+3x+9+2-5x}{\left(x-3\right)\left(x+3\right)}=\dfrac{5}{\left(x-3\right)\left(x+3\right)}\)

a: 2x-1=7

=>2x=8

=>x=4

Khi x=4 thì \(B=1-\dfrac{1}{4+3}=1-\dfrac{1}{7}=\dfrac{6}{7}\)

b: P=A:B

\(=\dfrac{2x^2+4x+13-3x-9-x^2+3x}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)

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

c: Để P là số nguyên thì \(x-3+5⋮x-3\)

=>\(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8\right\}\)

a: 2x-1=7

=>2x=8

=>x=4

Thay x=4 vào B, ta được:

\(B=1-\dfrac{1}{4+3}=1-\dfrac{1}{7}=\dfrac{6}{7}\)

b: \(A=\dfrac{2x^2+4x+13}{x^2-9}+\dfrac{3}{3-x}-\dfrac{x}{x+3}\)

\(=\dfrac{2x^2+4x+13-3x-9-x^2+3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+4x+4}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+2\right)^2}{\left(x-3\right)\left(x+3\right)}\)

b: P=A:B

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

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

c: Để P nguyên thì x-3+5 chia hết cho x-3

=>\(x-3\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{4;8\right\}\)

a: 2x-1=7

=>2x=8

=>x=4

Khi x=4 thì \(B=1-\dfrac{1}{4+3}=1-\dfrac{1}{7}=\dfrac{6}{7}\)

b: \(P=A:B\)

\(=\dfrac{2x^2+4x+13-3x-9-x^2+3x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+3-1}{x+3}\)

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

c: Để P nguyên thì x+2 chia hết cho x-3

=>x-3+5 chia hết cho x-3

=>\(x-3\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{4;8;-2\right\}\)

mik đag cần gấp mn giúp mik vs

đề bài bn ghi rõ ra dc ko ạ

(2 phần 3x)- (1 phần 2x-2)-(x-4 phần 6x-6x^2)

\(\dfrac{2}{3x}-\dfrac{1}{2x-2}-\dfrac{x-4}{6x-6x^2}\)

\(=\dfrac{2}{3x}-\dfrac{1}{2\left(x-1\right)}+\dfrac{x-4}{6x\left(x-1\right)}\)

\(=\dfrac{4\left(x-1\right)-3x+x-4}{6x\left(x-1\right)}=\dfrac{4x-4-2x-4}{6x\left(x-1\right)}\)

\(=\dfrac{2x-8}{6x\left(x-1\right)}=\dfrac{x-4}{3x\left(x-1\right)}\)

a: \(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}=\dfrac{-x^2+7x+78}{x\left(x+6\right)}=\dfrac{-\left(x-13\right)\left(x+6\right)}{x\left(x+6\right)}=\dfrac{-x+13}{x}\)

b: \(=\dfrac{x^3+1+x+1-x^2-2}{x^3+1}=\dfrac{x^3-x^2+x}{x^3+x}=\dfrac{x}{x+1}\)