\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)

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

\(x\left(x+1\right)+x\left(x-3\right)=4x\)

\(x^2+x+x^2-3x=4x\)

\(2x^2-2x=4x\)

\(2x^2-2x-4x=0\)

\(2x\left(x-3\right)=0\)

\(2x=0\Leftrightarrow x=0\)

hoặc 

\(x-3=0\Leftrightarrow x=3\)

b) \(ĐKXĐ:x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )

Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)

\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)

\(=\frac{-2}{x^2}\)

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

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

\(=x\left(x-3\right)\)

\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)

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

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

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

\(=\frac{x+3}{x+1}\)

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

a) \(\frac{3x+5}{2\left(x-1\right)}+\frac{4}{x-2}=\frac{\left(3x+5\right)\left(x-2\right)+4\cdot2\left(x-1\right)}{2\left(x-1\right)\left(x-2\right)}=\frac{3x^2-6x+5x-10+8x-8}{2\left(x-1\right)\left(x-2\right)}\)

\(=\frac{3x^2+7x-18}{2\left(x-1\right)\left(x-2\right)}\)

b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}=\frac{2x^2+1+4x\left(3-3x\right)+2\cdot3\left(1-2x\right)}{4x\left(1-2x\right)}=\frac{2x^2+1+12-12x+6-12x}{4x\left(1-2x\right)}\)\(=\frac{2x^2-24x+19}{4x\left(1-2x\right)}\)

Đề này... bạn xem lại đi. Chứ thế này thì dùng máy tính cũng không làm nổi T-T

mk ko biết làm 

xin lỗi bn nhae

xin lỗi vì đã ko giúp được bn

chcus bn học gioi!

nhae@@@

mình không biết làm

tk nhé@@@@@@@@@@@@@@@@@@@@

LOL

hihi

a) ... \(=\frac{3\left(2x-1\right)+2x\left(3x+3\right)+2x^2+1}{2x\left(2x-1\right)}=\frac{6x-3+6x^2+6x+2x^2+1}{2x\left(2x-1\right)}\)

\(=\frac{8x^2+12x-2}{2x\left(2x-1\right)}=\frac{4x^2+6x-1}{x\left(2x-1\right)}\)(hình như hết đơn giản được rồi, kết quả tạm vậy bạn nhé!)

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

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

c) ... \(=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)

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

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

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

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

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

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

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

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

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

\(=\frac{1}{x}\)

a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)

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

Chắc chắn đúng, mik nhaaaaaa

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

\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)

b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)

c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)

\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)

\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)

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

\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x-3y}{x\left(x+3y\right)}\)

e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

\(=\frac{3x-2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)

\(=\frac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)}-\frac{\left(3x-2\right)\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)

\(=\frac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-3x^3+6x^2-3x+2x^2-4x+2}{\left(x-1\right)^2\left(x+1\right)^2}\)

\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\) 

f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)

\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)