\(B=\left(\frac{21}{x^2-9}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)

\(B=\frac{2x^2-5x+12}{x^2-9}\cdot\frac{x+3}{x+2}\)

\(B=\frac{2x^2-5x-12}{\left(x-3\right)\left(x+2\right)}\)

\(B=\frac{2x^2-5x+12}{x^2-x-6}\)

Thik thì tách tiếp nha

đkxd: \(x\ne\left\{\pm3\right\}\)

a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)

=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)

=\(\frac{x+12}{x-3}\)

b)|2x+1|=5

<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2

với x=-3 thì B=\(\frac{-3}{2}\)

với x=2 thì B=-14

minh chua hieu buoc 1,2 của ban

1. \(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
2. Điều kiện \(x\ne3\) \(\Rightarrow\frac{-3}{5}=\frac{3}{x-3}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
3. \(B=\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

a) B=(\(\frac{21}{x^2-9}\)-\(\frac{x-4}{3-x}\)-\(\frac{x-1}{3+x}\)) : (1-\(\frac{1}{x+3}\)) (ĐK: x khác +-3)

=(\(\frac{21}{\left(x-3\right).\left(x+3\right)}\)+\(\frac{x-4}{x-3}\)-\(\frac{x-1}{x+3}\)) : (1-\(\frac{1}{x+3}\))

=(\(\frac{21+\left(x+4\right).\left(x+3\right)-\left(x-1\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}\):(\(\frac{x+3-1}{x+3}\))

=(\(\frac{3x+6}{\left(x-3\right).\left(x+3\right)}\)) . (\(\frac{x+3}{x+2}\))

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

=\(\frac{3}{x-3}\)

b) B=\(\frac{3}{x-3}\)=\(\frac{-3}{5}\)

(=) \(\frac{3.5}{x-3}\)=-3

(=) -3.(x-3) = 15

(=) -3x=6

(=) x=-2

vậy x=2 thì B=\(\frac{-3}{5}\)

c) B=\(\frac{3}{x-3}\)<0

(=) 3 < x - 3

(=) -x < - 3 - 3

(=) x > 6

Vậy với x > 6 thì B < 0

Ta có:

\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)

\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)

\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)

\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)

\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)

b/ Thế vô rồi tính nhé

Đoạn gần cuối thay y-x= 1 luôn 

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)

\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)

\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\)  giờ mới thay không biết đã tối giản chưa

a) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=x^2-2x+1-x^2+4=5-2x\)

mình nghĩ là câu b bạn ghi đề sai vì như thế không có hằng đẳng thức nhé

b)\(\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3=x^3-\frac{1}{27}-x^3+\frac{1}{27}+x^2-\frac{1}{3}x=x^2-\frac{1}{3}x\)

b,\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)

\(=\)\(\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)

\(=\)\(\left(x-\frac{1}{3}\right)\left(x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2}{3}x-\frac{1}{9}\right)\)

\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2}{3}x\right)\) \(=1+\frac{2}{3}x^2-\frac{1}{3x}-\frac{2}{9}x\)