Câu 2:

ĐKXĐ \(\hept{\begin{cases}x\ge0\\x-1\ne0\\x+2\sqrt{x}+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\\\left(\sqrt{x}+1\right)^2\ne0\end{cases}}\)

\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)

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

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

\(=\frac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)

\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}=\frac{2x}{x-1}\)

Câu 1 \(A=\sqrt{75}+1-3\sqrt{5}\)

\(C=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}=\dfrac{-3}{5}+\dfrac{3}{5}=0\)

= (3/6-2/6-1/6).(3/8+34/88-345/888)

= 0.(3/8+434/88-345/888)=0

      2.  8/3.2/5.3/8.10.19/92

= (8/3.3/8).(2/5.10).19/92

= 1.4.19/92

= 76/92

1) \(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)=\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)\)

                                                                                   \(=\left(\frac{1}{6}-\frac{1}{6}\right)\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)\)

                                                                                   \(=0\cdot\left(\frac{3}{8}+\frac{34}{88}+\frac{345}{888}\right)=0\)(số nào nhân với 0 cũng bằng 0)

2) \(\frac{8}{3}\cdot\frac{2}{5}\cdot\frac{3}{8}\cdot10\cdot\frac{19}{92}=\frac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\)

\(=\frac{2\cdot10\cdot19}{5\cdot92}=\frac{2\cdot2\cdot5\cdot19}{5\cdot2\cdot2\cdot23}=\frac{19}{23}\)

1. Có 1 thừa số là \(1-\frac{5}{5}=0\) nên tích sẽ bằng 0

2.\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{49}{50}=\frac{1}{50}\)

\(\frac{\left(2x^3+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

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

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

Thay x=\(\frac{1}{2}\)

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

\(=-1\)

\(\frac{-1}{3}+\frac{0,2-0,375+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)

\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)

\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{2}.\left(\frac{2}{10}-\frac{3}{8}+\frac{5}{11}\right)}\)

\(=\frac{-1}{3}+\frac{1}{-\frac{3}{2}}\)

\(=\frac{-1}{3}+\frac{-2}{3}=-\frac{3}{3}=-1\)