a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

= \(-33\sqrt{2}\)

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

= \(10-2\sqrt{21}+14\sqrt{21}\)

= \(10+12\sqrt{21}\)

Sửa đề: \(C=1+3^1+3^2+...+3^{100}\)

b) Ta có: \(C=1+3^1+3^2+...+3^{100}\)

\(\Leftrightarrow3\cdot C=3+3^2+...+3^{101}\)

\(\Leftrightarrow C-3\cdot C=1+3+3^2+...+3^{100}-3-3^2-...-3^{100}-3^{101}\)

\(\Leftrightarrow-2\cdot C=1-3^{101}\)

hay \(C=\dfrac{3^{101}-1}{2}\)

b) Ta có: C=1+31+32+...+3100C=1+31+32+...+3100

⇔3⋅C=3+32+...+3101⇔3⋅C=3+32+...+3101

⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101

⇔−2⋅C=1−3101

1, \(\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}=\left(\dfrac{2}{3}-\dfrac{3}{2}\right)\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{6}\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{8}+\dfrac{1}{2}=-\dfrac{1}{8}\)

2, \(\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=\left(-1\right)+1-\dfrac{3}{7}=0-\dfrac{3}{7}=-\dfrac{3}{7}\)

\(\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}=\dfrac{-5}{6}x\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{-5}{8}+\dfrac{1}{2}=\dfrac{-1}{8}\)

\(\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=-1+1-\dfrac{3}{7}=0-\dfrac{3}{7}=\dfrac{-3}{7}\)

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

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

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

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

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

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)

=\(\left(4-\dfrac{5}{12}\right).\dfrac{1}{2}+\dfrac{5}{24}\)

=\(4.\dfrac{1}{2}-\dfrac{5}{12}.\dfrac{1}{2}+\dfrac{5}{24}\)

=\(2-\dfrac{5}{24}+\dfrac{5}{24}=2\)

Giải:

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

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

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

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

\(=\dfrac{1}{3}.\left(-1\right)\) 

\(=\dfrac{-1}{3}\) 

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\) 

\(=\dfrac{43}{12}:2+\dfrac{5}{24}\) 

\(=\dfrac{43}{24}+\dfrac{5}{24}\) 

\(=2\)

\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)

\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)

\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)

\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)

\(\dfrac{-1}{6}\)

\(\dfrac{-1}{6}\)

\(\dfrac{-11}{15}\)

\(\dfrac{-9}{8}\)

1, \(\dfrac{-3}{5}:\dfrac{7}{5}-\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)

\(=\left(\dfrac{-3}{5}-\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)

\(=\dfrac{-6}{5}:\dfrac{7}{5}+\dfrac{13}{5}\)

\(=\dfrac{-6}{7}+\dfrac{13}{5}\\ =\dfrac{61}{35}\)

2, \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\left(-\dfrac{1}{9}\right)\)

\(=0,75-\dfrac{37}{12}+9\cdot\left(-\dfrac{1}{9}\right)\)

\(=-\dfrac{7}{3}+\left(-1\right)\\ -\dfrac{10}{3}\)

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

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

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1)

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

b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)

\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)

2)

a) \(2x+19=25\)

\(2x=25-19=6\)

\(x=3\)

b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)

\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)

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

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

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

\(\dfrac{\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{7}}{\dfrac{8}{3}-\dfrac{8}{5}+\dfrac{8}{7}}=\dfrac{2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}\right)}{8\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}\right)}\\ =\dfrac{2}{8}=\dfrac{1}{4}\)

2/3 - 2/5 + 2/7

= 70/105 - 42/105 + 20/105

= (70 - 42 + 20) / 105

= 48 / 105

= 16/35

8/3 - 8/5 + 8/7

= 280/105 - 168 / 105 + 120/105

= ( 280 - 168 + 120 ) / 105

= 232/105

2/3 - 2/5 + 2/7

= 70/105 - 42/105 + 30/105

= (70 - 42 + 30) / 105

= 58 / 105

a: Ta có: \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

=-5+2

=-3