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\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
a: \(A=\left(1+x+x^2-x\right):\dfrac{1-x^2}{x^3-x^2-x+1}\)
\(=\left(x^2+1\right)\cdot\dfrac{\left(x-1\right)\left(x^2-1\right)}{-\left(x^2-1\right)}=\left(1-x\right)\left(x^2+1\right)\)
b: Khi x=-5/3 thì \(A=\left(1+\dfrac{5}{3}\right)\left(\dfrac{25}{9}+1\right)=\dfrac{8}{3}\cdot\dfrac{34}{9}=\dfrac{272}{27}\)
c: Để A<0 thì 1-x<0
hay x>1
Mk có tâm rút gọn hộ bạn luôn rồi nè =))
a, ĐK : \(x\ne-2;3\)
b, \(A=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2}{x+2}\)
\(=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\frac{8-x+2x-6}{\left(x+2\right)\left(x-3\right)}\)
\(=\frac{x-2}{\left(x-2\right)\left(x-3\right)}=\frac{1}{x-3}\)
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)