a: \(3a^2bc^3\cdot\left(-2a^3b^5c\right)\cdot\left(-3a^5b^2c^2\right)\)

\(=3\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(a^3\cdot a^2\cdot a^5\right)\cdot\left(b\cdot b^5\cdot b^2\right)\cdot\left(c^3\cdot c\cdot c^2\right)\)

\(=18a^{10}b^8c^6\ge0\forall a,b,c\)

=>Ba đơn thức không thể cùng âm

\(-2a^5b^2;3a^2b^6\) cùng dấu

=>\(-2a^5b^2\cdot3a^2b^6>0\)

=>\(-6a^7b^8>0\)

=>\(6a^7b^8<0\)

=>\(a^7<0\)

=>a<0

b: E=0

=>\(\left(1-x\right)^4\cdot\left(-x\right)=0\)

=>\(\left[\begin{array}{l}1-x=0\\ x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=0\end{array}\right.\)

E>0

=>\(\left(1-x\right)^4\cdot\left(-x\right)>0\)

=>-x>0

=>x<0

E<0

=>\(\left(1-x\right)^4\cdot\left(-x\right)\) <0

=>-x<0

=>x>0

Đề thiếu. Bạn coi lại đề.

dạ không cần nữa đâu ạ 

1. b3+b= 3                                       

(b3+b)=3                            

b.(3+1)=3

b. 4= 3

b=\(\dfrac{3}{4}\)

a3+a= 3                                       b3

(a3+a)=3                            

a.(3+1)=3

a. 4= 3

a=\(\dfrac{3}{4}\)

2

a: Sửa đề: \(B=\left(\dfrac{2a}{a+3}+\dfrac{2}{3-a}+\dfrac{3}{a^2-9}\right):\dfrac{a+1}{a-3}\)

\(=\dfrac{2a^2-6a-2a-6+3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}=\dfrac{2a^2-8a-3}{\left(a+3\right)\left(a+1\right)}\)

b: |a|=2

=>a=2 hoặc a=-2

Khi a=2 thì \(B=\dfrac{2\cdot2^2-8\cdot2-3}{\left(2+3\right)\left(2+1\right)}=\dfrac{-11}{15}\)

Khi a=-2 thì \(B=\dfrac{2\cdot\left(-2\right)^2-8\cdot\left(-2\right)-3}{\left(-2+3\right)\left(-2+1\right)}=-21\)