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9 tháng 7 2016

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

Ta có:\(A=\frac{a^2}{a^2-1}-\frac{a^2}{1+a^2}.\frac{a}{a+1}-\frac{a^2}{1+a^2}.\frac{1}{a^2+a}\)

          \(A=\frac{a^2}{a^2-1}-\frac{a^3}{a+a^3+a^2+1}-\frac{a^2}{a+a^2+a^3+a^4}\)

27 tháng 6 2016

A= \(\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)

=\(-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)

B= \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)

b) ta có : A=\(-\frac{359}{390}\approx-0,9\)

B= \(\frac{13}{12}\approx1,08\)

=> A<x<B mà x nguyên => x=0 hoặc x=1

4 tháng 10 2021

Sai rồi

30 tháng 7 2016

\(\left(\frac{3a}{a^2-4}+\frac{1}{2-a}-\frac{2}{a+2}\right):\left(1-\frac{a^2+4}{a^2-4}\right)\)điều kiện : a khác {-2,2}

=\(\left(\frac{3a}{a^2-4}-\frac{a+2}{a^2-4}-\frac{2a-4}{a^2-4}\right):\left(-\frac{8}{a^2-4}\right)\)

=\(\left(\frac{3a-a-2-2a+4}{a^2-4}\right).\left(\frac{a^2-4}{-8}\right)\)

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

30 tháng 7 2016

\(=\left[\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{\left(a-2\right)}-\frac{2}{\left(a+2\right)}\right]:\left(\frac{a^2-4-a^2-4}{a^2-4}\right)=\left(\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}\right).\frac{\left(a-2\right)\left(a+2\right)}{-8}=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{-8}\)

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

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

\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x\right)^2-1}{2ax}\)

\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x+1\right)\left(a+x-1\right)}{2ax}\)

\(=\dfrac{\left(a+x+1\right)^2}{2ax}\)

26 tháng 6 2016

   \(\left(\frac{3a+1}{a^2-3a}+\frac{3a-1}{a^2+3a}\right)\):\(\frac{a^2+1}{a^2-9}\)

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

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

=\(\frac{3a^2+9a+a+3+3a^2-9a-a+3}{a\left(a-3\right)\left(a+3\right)}\): \(\frac{a^2+1}{\left(a-3\right)\left(a+3\right)}\)

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

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

=\(\frac{6}{a}\)

17 tháng 5 2016

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{399}{400}\Rightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{19.21}{20.20}\Rightarrow\frac{1.2.3...19}{2.3.4...20}.\frac{3.4.5...21}{2.3.4...20}\) \(\Rightarrow A=\frac{1}{20}.\frac{21}{2}=\frac{21}{40}\)

18 tháng 5 2016

Sửa đê, toán 6.

16 tháng 7 2016

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

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

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

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

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

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

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

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

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