=1/8x(3/2-1/4)+3/8x5/4

=1/8x5/4+3/8x5/4

=5/4x(1/8+3/8)

=5/8

\(C=\dfrac{1}{8}\times\dfrac{3}{2}-\dfrac{1}{8}\times\dfrac{1}{4}+\dfrac{3}{8}\times\dfrac{5}{8}\)

\(C=\dfrac{1}{8}\times\left(\dfrac{3}{2}-\dfrac{1}{4}\right)+\dfrac{3}{8}\times\dfrac{5}{4}\)

\(C=\dfrac{1}{8}\times\dfrac{5}{4}+\dfrac{3}{8}\times\dfrac{5}{4}\)

\(C=\left(\dfrac{1}{8}+\dfrac{3}{8}\right)\times\dfrac{5}{4}\)

\(C=\dfrac{1}{2}\times\dfrac{5}{4}\)

\(C=\dfrac{5}{8}\)

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

 60x [7/12+4/15]

60x153/180

=9180/180

b 1/2x2/3x3/4x4/5x5/6x6/7x7/8x8/9=40320/4032

\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)

\(\dfrac{3}{2}x-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\dfrac{3}{2}x=\dfrac{3}{5}+\dfrac{1}{5}\)

\(\dfrac{3}{2}x=\dfrac{4}{5}\)

   \(x=\dfrac{4}{5}:\dfrac{3}{2}\)

   \(x=\dfrac{4}{5}\cdot\dfrac{2}{3}\)

   \(x=\dfrac{8}{15}\)

\(\dfrac{1}{3}+x=\dfrac{3}{4}\)

        \(x=\dfrac{3}{4}-\dfrac{1}{3}\)

        \(x=\dfrac{9}{12}-\dfrac{4}{12}\)

        \(x=\dfrac{5}{12}\)

\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)

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

\(\dfrac{3}{2}x=\dfrac{1}{4}+\dfrac{2}{5}\)

\(\dfrac{3}{2}x=\dfrac{13}{20}\)

   \(x=\dfrac{13}{20}:\dfrac{3}{2}\)

   \(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)

   \(x=\dfrac{13}{30}\)

\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)

          \(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)

          \(\dfrac{3}{8}\cdot x=\dfrac{5}{4}\)

                \(x=\dfrac{5}{4}:\dfrac{3}{8}\)   

                \(x=\dfrac{5}{4}\cdot\dfrac{8}{3}\)

                \(x=\dfrac{10}{3}\)

`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`

`đk:x ne 0,x ne -2`

`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`

`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`

`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`

`=-4/(x+2)^2*(x+2)/4`

`=-1/(x+2)`

`b)P<0`

`<=>-1/(x+2)<0`

Vì `-1<0`

`<=>x+2>0`

`<=>x> -2`

`c)P=1/x+1(x ne 0)`

`<=>-1/(x+2)=1/x+1`

`<=>1/x+1+1/(x+2)=0``

`<=>x+2+x(x+2)+x=0`

`<=>x^2+4x+2=0`

`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\) 

`d)|2x-1|=3`

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\) 

`x=-1=>P=-1/(-1+2)=-1`

`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?

a) đk: \(x\ne-2;2\)

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

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

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

b) Để P < 0

<=> \(\dfrac{-1}{x+2}< 0\)

<=> x +2 > 0

<=> x > -2 ( x khác 2)

c) Để P= \(\dfrac{1}{x}+1\)

<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)

<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)

<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)

<=> x² + 4x + 2 = 0

<=> (x+2)² = 2

<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)

d) Để \(\left|2x-1\right|=3\)

<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)

Thay x = -1, ta có:

P = \(\dfrac{-1}{-1+2}=-1\)

a) ĐKXĐ: \(x\ne2;-2\)

\(P=\left(\dfrac{x}{x+2}-\dfrac{x^3-8}{x^3+8}.\dfrac{x^2-2x+4}{x^2-4}\right):\dfrac{4}{x+2}\)

\(=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4}{x+2}\)

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

\(=\left(\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right):\dfrac{4}{x+2}\)

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

b) \(P< 0\Rightarrow-\dfrac{1}{x+2}< 0\Rightarrow x+2>0\Rightarrow x>-2\)

\(\Rightarrow x>-2;x\ne2\)

c) \(P=\dfrac{1}{x}+1\Rightarrow\dfrac{-1}{x+2}=\dfrac{x+1}{x}\Rightarrow-x=\left(x+2\right)\left(x+1\right)\)

\(\Rightarrow-x=x^2+3x+2\Rightarrow x^2+4x+2=0\)

\(\Delta=4^2-2.4=8\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\end{matrix}\right.\)

d) \(\left|2x-1\right|=3\Rightarrow\left[{}\begin{matrix}2x-1=3\\1-2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}P=-\dfrac{1}{2+2}=-\dfrac{1}{4}\\P=-\dfrac{1}{-1+2}=-1\end{matrix}\right.\)

\(x-\dfrac{1}{2}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}+\dfrac{1}{2}\)

\(x=\dfrac{5}{4}\)

\(x+\dfrac{7}{8}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}-\dfrac{7}{8}\)

\(x=\dfrac{-1}{8}\)

\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)

\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)

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

Câu D ko bt

x=3/4-1/8

x=6/8-1/8

x=6/8

=2/5.(9/4-5/4)

=2/5.1

=2/5

\(x=\dfrac{3}{4}-\dfrac{1}{8}\)\(\)

\(x=\dfrac{6}{8}-\dfrac{1}{8}\)

\(x=\dfrac{5}{8}\)

\(=\dfrac{2}{5}x\left(\dfrac{9}{4}-\dfrac{5}{4}\right)\)

\(=\dfrac{2}{5}x1\)

\(=\dfrac{2}{5}\)

`x=3/4-1/8`

`x=5/8`

------------------------

`=2/5xx(9/4+5/4)`

`=2/5xx7/2=5/7`

\(\dfrac{1}{2}x.\dfrac{1}{4}x^2.\dfrac{x^3}{8}.2y.4y-8y^3=x.x^2.x^3.y.y.\dfrac{2.4}{2.4.8}-8y^3\\ =x^6.y^2.\dfrac{1}{8}-8y^3\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{8}\cdot x^3\cdot x^3\cdot8y^2-8y^3\)

\(=\dfrac{1}{8}x^6y^2-8y^3\)