B

Thay \(x=1;y=-1;z=3\) vào biểu thức ta có

\(1\cdot\left(-1\right)\cdot3+\dfrac{2\cdot1^2\cdot\left(-1\right)}{\left(-1\right)^2+1}\)

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

Đặt \(A=\left(\sqrt{2+\sqrt{2+\sqrt{2+...}}}\right)\)  nên \(A^2=2+\left(\sqrt{2+\sqrt{2+...}}\right)\) ( có vô hạn dấu căn)

hay \(A^2=2+A\Leftrightarrow A^2-A-2=0\Leftrightarrow\left(A+1\right)\left(A-2\right)=0\)

Vì A>0 nên A=2

tick nha 

1) \(\dfrac{17}{5}\cdot\dfrac{-31}{125}\cdot\dfrac{1}{2}\cdot\dfrac{10}{17}\cdot\dfrac{-1}{2^3}\)

\(=\dfrac{17\cdot\left(-31\right)\cdot1\cdot2\cdot5\cdot\left(-1\right)}{5\cdot125\cdot2\cdot17\cdot8}\)

\(=\dfrac{\left(-31\right)\left(-1\right)}{125\cdot8}\\ =\dfrac{31}{1000}\)

3) \(58\left(3\dfrac{1}{29}-2\dfrac{1}{58}\right):\dfrac{1}{3}\)

\(=58\left(\dfrac{88}{29}-\dfrac{117}{58}\right)\cdot3\\ =\left(58\cdot\dfrac{88}{29}-58\cdot\dfrac{117}{58}\right)\cdot3\)

\(=\left(176-117\right)\cdot3\\ =59\cdot3\\ =177\)

4) \(A=1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\cdot1\dfrac{1}{5}\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}\\ =\dfrac{3\cdot4\cdot5\cdot6}{2\cdot3\cdot4\cdot5}\\ =\dfrac{6}{2}\\ =3\)

\(x\left(\dfrac{124124}{125125}-1\right)=\dfrac{2222}{5555}:\left[\left(\dfrac{2010}{2121}-1\right):\left(\dfrac{4040}{4141}-1\right)\right]\)

\(x\left(\dfrac{124}{125}-1\right)=\dfrac{2}{5}:\left(-\dfrac{37}{707}:\dfrac{-1}{41}\right)\)

\(-\dfrac{1}{125}x=\dfrac{2}{5}:\dfrac{1517}{707}\)

\(-\dfrac{1}{125}x=\dfrac{1414}{7585}\)

\(x=\dfrac{1414}{7585}:\dfrac{-1}{125}\\ x=\dfrac{-35350}{1517}\)

2) \(\left(\dfrac{1}{6}+2\dfrac{1}{2}-10,75\right)\cdot x-7=\left(\dfrac{2}{5}+\dfrac{3}{8}+0,025\right):0,1\)

\(\left(\dfrac{1}{6}+\dfrac{5}{2}-\dfrac{43}{4}\right)x-7=\left(\dfrac{2}{5}+\dfrac{3}{8}+\dfrac{1}{40}\right):\dfrac{1}{10}\)

\(-\dfrac{97}{12}x-7=\dfrac{4}{5}\cdot10\)

\(-\dfrac{97}{12}x=8+7\\ -\dfrac{97}{12}x=15\\ x=15:\left(-\dfrac{97}{12}\right)\\ x=-\dfrac{180}{97}\)

a)Sắp xếp : \(f\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x\)

\(g\left(x\right)=-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}x\)

Ta có : \(f\left(x\right)+g\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}x\)

\(=12x^4-11x^3+2x^2-\dfrac{1}{2}x\)

\(f\left(x\right)-g\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x+x^5-5x^4+2x^3-4x^2+\dfrac{1}{4}x\)

\(=2x^5+2x^4-7x^3-6x^2\)

a) \(1\dfrac{1}{5}+\dfrac{5}{9}+\dfrac{4}{5}+\dfrac{4}{9}\)

\(=\left(\dfrac{6}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\\ =2+1\\ =3\)

b) \(2\dfrac{-7}{10}:\left(\dfrac{5}{7}+\dfrac{3}{14}\right)\)

\(=-\dfrac{27}{10}:\dfrac{13}{14}\\ =-\dfrac{27}{10}\cdot\dfrac{14}{13}\\ =-\dfrac{189}{65}\)

c) \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\\ =\dfrac{7}{30}\)

Gọi số hs là x

Ta có: \(x\cdot\dfrac{3}{7}=18\)

Vậy\(x=18:\dfrac{3}{7}=42\)

Đáp số:42 hs