a) \(M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)

\(\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)

\(\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)

\(\Leftrightarrow M=\dfrac{2}{1-2x}\)

b) \(M=\dfrac{2}{1-2x}=\dfrac{-2}{3}\)

\(\Rightarrow2.3=\left(1-2x\right).\left(-2\right)\)

\(\Rightarrow6=-2+4x\)

\(\Rightarrow4x=6-\left(-2\right)\)

\(\Rightarrow4x=6+2\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=8:4\)

\(\Rightarrow x=2\)

Vậy \(M=\dfrac{-2}{3}\) thì \(x=2\)

c) Để \(M=\dfrac{2}{1-2x}\in Z\) \(\Leftrightarrow2⋮1-2x\)

\(\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{1;0\right\}\)

Vậy \(x=1\) hoặc \(x=0\) thì \(M\in Z\)

a) M = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

= \(\left(\dfrac{1}{1-x}+\dfrac{2}{1+x}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{x^2-1}{1-2x}\)

= \(\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

= \(\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)\(=\dfrac{-2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

= \(\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

=\(\dfrac{2}{1-2x}\)

b) M = \(\dfrac{-2}{3}\Leftrightarrow\dfrac{2}{1-2x}=\dfrac{-2}{3}\)

=> 2 . 3 = -2 (1 - 2x) (tích chéo)

=> 6 = -2 + 4x

=> 6 + 2 - 4x = 0

=> 8 - 4x = 0

=> 4x = 8

=> x = 2 (thỏa mãn đkxđ)

Vậy để M = \(\dfrac{-2}{3}\) thì x = 2

\(a.\)

\(P=\left[\left(\dfrac{1}{x^2}+1\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)\right].\dfrac{x-1}{x^3}\)

\(P=\left[\left(\dfrac{1}{x^2}+\dfrac{x^2}{x^2}\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+\dfrac{x}{x}\right)\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^2}.\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{x+1}{x}\right)\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^2\left(x^2+2x+1\right)}+\dfrac{2}{x\left(x+1\right)^2}\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2}{x^3+2x^2+x}\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x\left(x^3+2x^2+x\right)}\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x^4+2x^3+x^2}\right].\dfrac{x-1}{x^3}\)

\(P=\dfrac{x^2+1+2x}{x^4+2x^3+x^2}.\dfrac{x-1}{x^3}\)

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

\(P=\dfrac{1}{x^2}.\dfrac{x-1}{x^3}\)

\(P=\dfrac{x-1}{x^5}\)

Làm nốt đi cậu ! Bạn ko làm là tớ làm đó @@

- A =\(\frac{x^2+3+2x-6-x-3}{x^2-9}\) 

- A =\(\frac{x^2+x-6}{x^2-9}\)

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

- A = \(\frac{x-2}{x-3}\)

a)

\(4x-10< 0\\ 4x< 10\\ x< \dfrac{10}{4}=\dfrac{5}{2}\)

b)

\(2x+x+12\ge0\\ 3x\ge-12\\ x\ge-\dfrac{12}{3}=-4\)

c)

\(x-5\ge3-x\\ 2x\ge8\\ x\ge4\)

d)

\(7-3x>9-x\\ -2>2x\\ x< -1\)

đ)

\(2x-\left(3-5x\right)\le4\left(x+3\right)\\ 2x-3+5x\le4x+12\\ 3x\le15\\ x\le5\)

e)

\(3x-6+x< 9-x\\ 5x< 15\\ x< 3\)

f)

\(2t-3+5t\ge4t+12\\ 3t\ge15\\ t\ge5\)

g)

\(3y-2\le2y-3\\ y\le-1\)

h)

\(3-4x+24+6x\ge x+27+3x\\ 0\ge2x\\ 0\ge x\)

i)

\(5-\left(6-x\right)\le4\left(3-2x\right)\\ 5-6+x\le12-8x\\ \\ 9x\le13\\ x\le\dfrac{13}{9}\)

k)

\(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\ 10x-15-20x+28\ge19-2x-22\\ 13-10x\ge-2x-3\\ -8x\ge-16\\ x\le\dfrac{-16}{-8}=2\)

l)

\(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\ \dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\ \Rightarrow40x-100-90x+30< 36-12x-30x+15\\ 130-50x< 51-42x\\ 92x< -79\\ x< -\dfrac{79}{92}\)

m)

\(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\ \dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\ \Rightarrow10x-3+2x>7x-5+2x\\ 12x-3>9x-5\\ 3x>-2\\ x>-\dfrac{2}{3}\)

n)

\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\ \dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\ \Rightarrow28x-8-24x< 60-3x+6\\ 4x-8< -3x+66\\ 7x< 74\\ x< \dfrac{74}{7}\)

a) \(4x-10< 0\)

\(\Leftrightarrow4x< 10\)

\(\Leftrightarrow x< \dfrac{5}{2}\)

b) ???

c) \(x-5\ge3-x\)

\(\Leftrightarrow2x-5\ge3\)

\(\Leftrightarrow2x\ge8\)

\(\Leftrightarrow x\ge4\)

d) \(7-3x>9-x\)

\(\Leftrightarrow7-2x>9\)

\(\Leftrightarrow-2x>2\)

\(\Leftrightarrow x< -1\)

đ) ???

e) \(3x-6+x< 9-x\)

\(\Leftrightarrow4x-6< 9-x\)

\(\Leftrightarrow5x-6< 9\)

\(\Leftrightarrow5x< 15\)

\(\Leftrightarrow x< 3\)

f) ???

g) ???

h) \(3-4x+24+6x\ge x+27+3x\)

\(\Leftrightarrow2x+27\ge4x+27\)

\(\Leftrightarrow-2x\ge0\)

\(\Leftrightarrow x\le0\)

i) \(5-\left(6-x\right)\le4\left(3-2x\right)\)

\(\Leftrightarrow5-6+x\le12-8x\)

\(\Leftrightarrow x-1\le12-8x\)

\(\Leftrightarrow9x-1\le12\)

\(\Leftrightarrow9x\le13\)

\(\Leftrightarrow x\le\dfrac{13}{9}\)

k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28\ge19-2x-22\)

\(\Leftrightarrow-10x+23\ge-3-2x\)

\(\Leftrightarrow-8x+13\ge-3\)

\(\Leftrightarrow-8x\ge-16\)

\(\Leftrightarrow x\ge2\)

l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)

\(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\)

\(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\)

\(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\)

\(\Leftrightarrow x>-\dfrac{121}{8}\)

m, n) làm tương tự:

đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\)

a)

\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)