a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)

\(=6x^4-13x^3+19x^2-17x+5\)

b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)

\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)

c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)

\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)

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

\(\Rightarrow A=3x\)

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

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

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

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

\(\Leftrightarrow1=\dfrac{A}{\left(x+2\right)}\)

\(\Leftrightarrow A=x+2\)

Bài 1:

\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)

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

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

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

Bài 2.

Sửa đề

a) \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)

Giải

Ta sẽ phân tích vế phải

VP = \(\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)

VP = \(\dfrac{a\left(x^2-4\right)+bx\left(x+2\right)+cx\left(x-2\right)}{x\left(x^2-4\right)}\)

VP = \(\dfrac{ax^2-4a+bx^2+2bx+cx^2-2cx}{x\left(x^2-4\right)}\)

VP = \(\dfrac{x^2\left(a+b+c\right)+2x\left(b-c\right)-4a}{x\left(x^2-4\right)}\)

Tương tự , ta cũng sẽ phân tích VT

VT = \(\dfrac{2x.5-4}{x\left(x^2-4\right)}\)

Đồng nhất hai VT và VP , ta có :

\(x^2\left(a+b+c\right)+2x\left(b-c\right)-4a=2.5x-4\)

* a + b + c = 0 => 1 + c + 5 + c = 0 => 2c = - 6 => c = - 3

* b - c = 5 => b = c + 5 => b = - 3 + 5 => b = 2

* a = 1

Vậy , a = 1 ; b = 2 ; c = -3

b) Ta sẽ phân tích VP

VP = \(\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)

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

VP = \(\dfrac{ax^2+ax+a+bx^2-bx+cx-c}{x^3-1}\)

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

Đồng nhất VP và VT , ta được :

\(x^2\left(a+b\right)+x\left(a-b+c\right)+a-c=1\)

* a + b = 0 => a = - b => b = \(-\dfrac{1}{3}\)

* a - b + c = 0 => a + a + a - 1 = 0 => 3a = 1 => a = \(\dfrac{1}{3}\)

* a - c = 1 => c = a - 1 => c = \(\dfrac{1}{3}\) - 1 = \(-\dfrac{2}{3}\)

Vậy , a = \(\dfrac{1}{3}\) ; b = \(-\dfrac{1}{3}\); c = \(-\dfrac{2}{3}\)

Bài 1 bạn Giang làm rồi thì thôi nhé

\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

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

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

\(=x^2-2x-5\)

\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)

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

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

\(=2x-3\)

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

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

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

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

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

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)