a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)

Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)

Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)

- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)

Vậy ...

b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;

đúng rồi

 chó điên

\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)

\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)

\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)

\(A=\sqrt{\left(a-3\right)^2}-3a\)

=3-a-3a

=3-4a

2a - (5- 4a) +(6a -1) -(2+a)

= -10a - 8a^2 +6a -1 -2 -a

= -8a^2 -5a -3

5a - 2b +3 - (2a -5b +6) +(a+3b -1)

= 5a -2b +3 -2a+5b -6 +a + 3b -1

= 4a +6b -4

6x(x-1) -1(6x^2 -8x +3) = 7 -(x-1)

6x^2 -6x - 6x^2 + 8x -3 = 7 -x +1

3x = 11

x= 11/3

7x(2x-1) - (14x^2 -8x +5) = 7- (-2x +3)

14x^2 - 7x - 14x^2 + 8x - 5 = 7 + 2x -3

-x = 9

x=-9  

a) = \(12a^2b\left(a^2-b^2\right)\)

= \(12a^4b-12a^2b^3\)

b)nhân ra :

= \(2x^4-16x^3+4x^2-3x^3+24x^2-6x+5x^2-40x+10\)

= \(2x^4-19x^3+33x^2-46x+10\)

Tìm x:

a) \(\frac{1}{4}x^2-\left(\frac{1}{4}x^2-2x\right)=-14\)

= \(\frac{1}{4}x^2-\frac{1}{4}x^2+2x=-14\)

=\(2x=-14=>x=-7\)

b) \(x^3+27-x\left(x^2-1\right)=27\)

= \(x^3+27-x^3+x=27\)

= \(27+x=27=>x=0\)