\(\left(\dfrac{1}{2}\right)^2.\left(\dfrac{2}{3}\right)^2.\left(\dfrac{3}{4}\right)^2.....\left(\dfrac{9}{10}\right)^2\)

\(=\left(\dfrac{1}{1}\right)^2.\left(\dfrac{1}{1}\right)^2.\left(\dfrac{1}{1}\right)^2....\left(\dfrac{1}{10}\right)^2\)

\(=\dfrac{1^2}{10^2}=\dfrac{1}{100}\)

\(\left(\dfrac{1}{2}\right)^2.\left(\dfrac{2}{3}\right)^2.\left(\dfrac{3}{4}\right)^2...\left(\dfrac{9}{10}\right)^2\)

\(=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right)^2=\left(\dfrac{1}{10}\right)^2=\dfrac{1}{100}\)

a: \(A=\left(\dfrac{6x+4}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)

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

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

b: Để A là số nguyên thì \(3x-2\sqrt{3x}+1⋮\sqrt{3x}-2\)

=>\(\sqrt{3x}-2\in\left\{1;-1;3;-3\right\}\)

=>\(3x\in\left\{9;1;25\right\}\)

hay x=3

ai giup mik dc ko ak pls mik can gap

\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)

a. ĐKXĐ : x>1.

b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)

c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:

\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)

Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).

a)ĐKXĐ:x>0

P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)

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

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

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

=\(\frac{4-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{1}\)

=\(\frac{4-\sqrt{x}}{\sqrt{x}-1}\)

b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)

\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)

\(\Leftrightarrow21-9\sqrt{x}=0\)

\(\Leftrightarrow-9\sqrt{x}=-21\)

\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{21}{9}\)

Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)

c)Còn phần c thì mik chịu

...

a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)

\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)

\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{119}{720}\)

b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)

\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)

\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)

\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)

\(B=1\)

a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)

   A  = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

   A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

   A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)

  A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)

 A = \(\dfrac{119}{720}\)

b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]

    B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]

    B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]

   B  = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))

B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)

B = 1

\(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)

\(A=\left(-\dfrac{2}{4}-\dfrac{1}{4}\right).\left(-\dfrac{1}{5}\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{3}\right)\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{1}{5}\right)+\dfrac{1}{48}-\dfrac{1}{18}\)

\(A=\dfrac{3}{20}+\dfrac{1}{48}-\dfrac{1}{18}=\dfrac{108}{720}+\dfrac{15}{720}-\dfrac{40}{720}=\dfrac{83}{720}\)

\(P=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3+x\right)\left(2x+3-x\right)}{3\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{3\left(x+1\right)\left(x+3\right)}{3\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{5\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{\left(x+5\right)}{x+1}-\dfrac{x+1}{x+5}\)

\(=\dfrac{5x^2+30x+45+x^2+10x+25-x^2-2x-1}{\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{5x^2+38x+69}{\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{5x^2+38x+69}{x^2+6x+5}\)

Để P là số nguyên thì \(5x^2+30x+25+8x+34⋮x^2+6x+5\)

=>\(8x+34⋮x^2+6x+5\)

=>\(\left\{{}\begin{matrix}8x+34⋮x+1\\8x+34⋮x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+8+26⋮x+1\\8x+40-6⋮x+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+1\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\\x+5\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\end{matrix}\right.\)

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