a)  \(ĐKXĐ:\) \(x\ne\pm1\)

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

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

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

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

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

Nhanh , mọi người , mình đang cần gấp lắm

        \(\frac{3x-4}{5}+\frac{x}{6}=\frac{x+1}{2}\)

\(\Leftrightarrow\)\(\frac{6\left(3x-4\right)}{30}+\frac{5x}{30}=\frac{15\left(x+1\right)}{30}\)

\(\Rightarrow\)\(6\left(3x-4\right)+5x=15\left(x+1\right)\)

\(\Leftrightarrow\)\(18x-24+5x=15x+15\)

\(\Leftrightarrow\)\(8x=39\)

\(\Leftrightarrow\)\(x=\frac{39}{8}\)

Vậy...

a)   Ta có:   \(\frac{4}{8}=\frac{5}{10}=\frac{6}{12}\left(=\frac{1}{2}\right)\)

hay   \(\frac{AB}{A'B'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}\)

\(\Rightarrow\)\(\Delta A'B'C'~\Delta ABC\) 

b)   \(\Delta A'B'C'~\Delta ABC\)

\(\Rightarrow\)\(\frac{P_{A'B'C'}}{P_{ABC}}=\frac{A'B'}{AB}=\frac{8}{4}=2\)

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đừng đăng linh tinh nha bạn

\(a)\) Ta có : 

\(M=\frac{2\left|x-3\right|}{x^2+2x-15}=\frac{2\left|x-3\right|}{\left(x^2+2x+1\right)-16}=\frac{2\left|x-3\right|}{\left(x+1\right)^2-16}=\frac{2\left|x-3\right|}{\left(x+1\right)^2-4^2}=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}\)

+) Nếu \(x-3\ge0\) \(\Rightarrow\) \(x\ge3\) ta có : 

\(M=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}=\frac{2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\frac{2}{x+5}\)

+) Nếu \(x-3< 0\)\(\Rightarrow\)\(x< 3\) ta có : 

\(M=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}=\frac{-2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\frac{-2}{x+5}\)

Vậy : +) Nếu \(x\ge3\) thì \(M=\frac{2}{x+5}\) 

         +) Nếu \(x< 3\) thì \(M=\frac{-2}{x+5}\)

Chúc bạn học tốt ~ 

1 + x² là mẫu số của phân số thứ nhất nha

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

\(A=\frac{x\left(1-x\right)^2.\left(1+x\right)^2}{1+x^2}:\left\{\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right].\left[\frac{\left(1+x\right)\left(1-x+x^2\right)}{1+x}-x\right]\right\}\)

\(A=\frac{x.\left(1-x\right)^2.\left(1+x\right)^2}{1+x^2}:\left[\left(1+2x+x^2\right).\left(1-2x+x^2\right)\right]\)

\(A=\frac{x.\left(1-x\right)^2.\left(1+x\right)^2}{1+x^2}:\left[\left(1-x\right)^2.\left(1+x\right)^2\right]\)

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

\(A=\frac{x}{1+x^2}\)

dgjc gfjd jdYID jĐUUKhd DKKKZ 

help me

\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
(x+4)/5 - (x+4*5)/5 = x/3 - (x-2)/2
(6x+24)/5 = 2x/6 - (3x-6)/6
6x/5+24/5 = 2x/6 - 3x/6 + 6/6
6x/5+24/5 = -x/6 + 1
6x/5+x/6 = 1 - 24/5
41x/30 = -19/5
41x = -19/5*30
41x = -114
x = -114/41
Vậy x = \(\frac{-114}{41}\)
P/S: những dấu "/" là dấu gạch của phân số, vì lười quá nên ko viết hẳn ra