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27 tháng 11 2017

a) Tớ làm luôn nhé , không chép lại đề đâu

P = \(\left[\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right].\dfrac{x\left(x+6\right)}{2x-6}\)

ĐKXĐ : x # -6 ; x # 6 ; x # 0 ; x # 3 . Khi đó , ta có :

P = \(\left[\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]\).\(\dfrac{x\left(x+6\right)}{2x-6}\)

P = \(\dfrac{x^2-x^2+12x-36}{x-6}.\dfrac{1}{2x-6}\)

P = \(\dfrac{6\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}=\dfrac{6}{x-6}\)

b) Tương tự

\(A=\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}-\dfrac{x}{x-6}\)

\(=\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{12\left(x-3\right)}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(=\dfrac{12}{2\left(x-6\right)}-\dfrac{x}{x-6}=\dfrac{6-x}{x-6}=-1\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a) 4x2(5x2 + 3) – 6x(3x3 – 2x + 1) – 5x3 (2x – 1)

= 4x2 . 5x2 + 4x2 . 3 – [6x . 3x3 + 6x . (-2x) + 6x . 1] – [5x3 . 2x + 5x3 . (-1)]

= 20x4 + 12x2 – (18x4 – 12x2 + 6x) – (10x4 – 5x3)

= 20x4 + 12x2 - 18x4 + 12x2 - 6x - 10x4 + 5x3

= (20x4 – 18x4 - 10x4 ) + 5x3 + (12x2 + 12x2 ) – 6x

= -8x4 + 5x3 + 24x2 – 6x

\(\begin{array}{l}b)\dfrac{3}{2}x\left( {{x^2} - \dfrac{2}{3}x + 2} \right) - \dfrac{5}{3}{x^2}(x + \dfrac{6}{5})\\ = \dfrac{3}{2}x.{x^2} + \dfrac{3}{2}x.( - \dfrac{2}{3}x) + \dfrac{3}{2}x.2 - (\dfrac{5}{3}{x^2}.x + \dfrac{5}{3}{x^2}.\dfrac{6}{5})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - (\dfrac{5}{3}{x^3} + 2{x^2})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - \dfrac{5}{3}{x^3} - 2{x^2}\\ = (\dfrac{3}{2}{x^3} - \dfrac{5}{3}{x^3}) + ( - {x^2} - 2{x^2}) + 3x\\ = \dfrac{{ - 1}}{6}{x^3} - 3{x^2} + 3x\end{array}\)

24 tháng 5 2022

Với `x \ne -5,x \ne -1` có:

`A=[x+2]/[x+5]+[-5x-1]/[x^2+6x+5]-1/[1+x]`

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+5)(x+1)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+5)(x+1)]`

`A=[x^2-3x-4]/[(x+5)(x+1)]`

`A=[(x-4)(x+1)]/[(x+5)(x+1)]`

`A=[x-4]/[x+5]`

24 tháng 5 2022

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

AH
Akai Haruma
Giáo viên
5 tháng 4 2018

Lời giải:

ĐKXĐ: \(x\neq -3; x\neq \pm 6; x\neq 0\)

Ta có:

\(A=\left(\frac{x}{x^2-36}-\frac{x+6}{x^2-6x}\right): \frac{2x+6}{x^2-6x}-\frac{x}{x+6}\)

\(A=\left(\frac{x}{x^2-36}-\frac{x+6}{x^2-6x}\right).\frac{x^2-6x}{2x+6}-\frac{x}{x+6}\)

\(=\frac{x(x^2-6x)}{(x^2-36)(2x+6)}-\frac{(x+6)(x^2-6x)}{x^2-6x)(2x+6)}-\frac{x}{x+6}\)

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

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

\(=\frac{x^2-(x+6)^2-x(2x+6)}{(x+6)(2x+6)}=\frac{-(2x^2+18x+36)}{2x^2+18x+36}=-1\)

Bài 2:

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

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

7 tháng 10 2023

a) ĐKXĐ: \(x\ne0;x\ne6;x\ne-6\)

b) \(A=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}+\dfrac{x}{6-x}\)

\(A=\left[\dfrac{x}{\left(x+6\right)\left(x-6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right]:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\left[\dfrac{x^2}{x\left(x+6\right)\left(x-6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right]:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{x^2-x^2+12x-36}{x\left(x+6\right)\left(x-6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{12x-36}{x\left(x+6\right)\left(x-6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}+\dfrac{x}{6-x}\)

\(A=\dfrac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}:\dfrac{2\left(x-3\right)}{x\left(x+6\right)}-\dfrac{x}{x-6}\)

\(A=\dfrac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}-\dfrac{x}{x-6}\)

\(A=\dfrac{6}{x-6}-\dfrac{x}{x-6}\)

\(A=\dfrac{6-x}{x-6}\)

\(A=-\dfrac{x-6}{x-6}\)

\(A=-1\)

Vậy giá trị của A không phụ thuộc vào giá trị của biến 

28 tháng 6 2017

Phép trừ các phân thức đại số

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)