a) `A=a. 1/3 + a. 1/4 - a.1/6 = a. (1/3+1/4 -1/6)=a. 5/12`

Thay `a=-3/5: A=-3/5 . 5/12 =-1/4`

b) `B=b. 5/6+ b. 3/4-b. 1/2=b.(5/6+3/4-1/2)=b. 13/12`

Thay `b=12/13: B=12/13 . 13/12=1`.

a) Ta có: \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\)

\(=a\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\)

\(=a\cdot\left(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{2}{12}\right)\)

\(=a\cdot\dfrac{5}{12}\)

\(=\dfrac{-3}{5}\cdot\dfrac{5}{12}=\dfrac{-1}{4}\)

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

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

\(=b\cdot\left(\dfrac{10}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\)

\(=b\cdot\dfrac{5}{4}\)

\(=\dfrac{12}{13}\cdot\dfrac{5}{4}=\dfrac{60}{52}=\dfrac{15}{13}\)

a) \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\\ A=a\cdot\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\\ A=a\cdot\dfrac{-5}{12}\)

Khi \(a=\dfrac{-3}{5}\), ta có:

\(A=\dfrac{-3}{5}\cdot\dfrac{-5}{12}\\ A=\dfrac{1}{4}\)

Vậy khi \(a=\dfrac{-3}{5}\) thì \(A=\dfrac{1}{4}\)

b. \(B=b\cdot\dfrac{5}{6}+b\cdot\dfrac{3}{4}-b\cdot\dfrac{1}{2}\\ B=b\cdot\left(\dfrac{5}{6}+\dfrac{3}{4}-\dfrac{1}{2}\right)\\ B=b\cdot\dfrac{13}{12}\)

Khi \(a=\dfrac{12}{13}\), ta có:

\(B=\dfrac{12}{13}\cdot\dfrac{13}{12}\\ B=1\)

Vậy khi \(a=\dfrac{-3}{5}\) thì B = 1

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\(A=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\\ B=\dfrac{25}{11}\times\dfrac{13}{12}\times\dfrac{-11}{5}=\dfrac{5\times13\times\left(-1\right)}{1\times12\times1}=\dfrac{-65}{12}\\ C=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\times\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\times\dfrac{-2}{5}=\dfrac{-11}{50}\)

\(B< -1< C< 0< A\\ \Leftrightarrow B< C< A\)

a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)

b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

a.=\(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{61}{4}:\dfrac{7}{12}=\dfrac{471}{14}-\dfrac{183}{7}=\dfrac{15}{2}\)

b.=\(\dfrac{2}{15}-\dfrac{2}{3}+\dfrac{1}{5}=-\dfrac{1}{3}\)

c.\(\left(\dfrac{10}{3}+2.5\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}=\dfrac{35}{6}:\left(-\dfrac{31}{30}\right)-\dfrac{11}{31}=-\dfrac{175}{31}-\dfrac{11}{31}=-6\)

d.\(\left[6+\dfrac{1}{8}-\dfrac{1}{2}\right]:\dfrac{3}{12}=\dfrac{45}{8}:\dfrac{3}{12}=\dfrac{45}{2}\)

\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\dfrac{\sqrt{3}-1}{4}\right)^2}=\dfrac{\sqrt{12+2\sqrt{3}}}{4}\)

\(\Rightarrow2\cos\alpha=\dfrac{\sqrt{12+2\sqrt{3}}}{2}\). Chọn B.

a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)

\(=\dfrac{1}{2016}\)

b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)

\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)

hay x=10

Vậy: x=10

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

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

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

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

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

b: \(x^2+x=12\)

=>\(x^2+x-12=0\)

=>(x+4)(x-3)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Thay x=3 vào A, ta được:

\(A=\dfrac{\left(3+1\right)^2}{3}=\dfrac{16}{3}\)

Khi x=-4 thì \(A=\dfrac{\left(-4+1\right)^2}{-4}=\dfrac{9}{-4}=-\dfrac{9}{4}\)

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

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

\(=\dfrac{x^2+2x+1-4x}{x}=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}\)>0 với mọi x>0

=>A>4

a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)

\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)

a: TXĐ: D=[0;+\(\infty\))\{1}

Ta có: \(P=\left(\dfrac{3}{\sqrt{x}+1}-\dfrac{1}{x-1}\right):\dfrac{1}{\sqrt{x}+1}\)

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

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