\(A=\frac{x+3a}{2-x}+\frac{x-3a}{2+x}-\frac{2a}{4-x^2}+a\)

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

\(=\frac{2x+x^2+6a+3ax+2x-x^2-6a+3ax-2a}{\left(2-x\right)\left(2+x\right)}+a\)

\(=\frac{6ax+4x-2a+a\left(4-x^2\right)}{\left(2-x\right)\left(2+x\right)}=\frac{6ax+4x-2a+4a-a\cdot x^2}{\left(2-x\right)\left(2+x\right)}=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)

\(-a\cdot x^2+6a\cdot x+2a\)

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

\(=\frac{-a^3+18a^3+12a^2+2a\left(9a^2+12a+4\right)}{\left(3a+2\right)^2}\)

\(=\frac{17a^3+12a^2+18a^3+24a^2+8a}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}\)

\(\left(2-x\right)\left(2+x\right)\)

\(=\left(2-\frac{a}{3a+2}\right)\left(2+\frac{a}{3a+2}\right)=\frac{6a+4-a}{\left(3a+2\right)}\cdot\frac{6a+4+a}{3a+2}\)

\(=\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}\)

\(A=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)

\(=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}:\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(5a+4\right)\left(7a+4\right)}\)

Bài 2:

a) \(\left(x+5\right)^2=x^2+10x+25\)

b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)

c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)

d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)

e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)

f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)

Bài 1:

$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$

$=4a.2b=8ab$

$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$

$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$

$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$

$=m^2+2mn+n^2=(m+n)^2$

Bài 2: 

3x + 2(5 - x) = 0

<=> 3x + 10 - 2x = 0

<=> x + 10 = 0

<=> x = 0 - 10

<=> x = -10

=> x = -10

Bài 3: 

6(3q + 4q) - 8(5p - q) + (p - q)

= 6.3p + 6.4q - 8.5p - (-8).q + p - q

= 18p + 24q - 40p + 8q + p - q

= (18p - 40p + p) + (24q + 8q - q)

= -21p + 31q

cho mik hỏi là dấu căn bậc 2 là của mik số 4 hay là cả 4a^6 và các phần kia cug vậy

https://hoc24.vn/hoi-dap/question/177629.html (câu 2)