1) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)

\(=-6\sqrt{2}\)

2) \(\sqrt{50}-\sqrt{18}+\sqrt{200}-\sqrt{162}\)

\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}\)

\(=3\sqrt{2}\)

3) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)

\(=5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)

\(=-2\sqrt{5}\)

4) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)

\(=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)

\(=4\sqrt{3}\)

5) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{3}\sqrt{3}\)

\(=-\dfrac{17}{3}\sqrt{3}\)

a: \(=\dfrac{-1}{4}+\dfrac{7}{33}-\dfrac{5}{3}-\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{48}{49}\)

\(=\dfrac{-3}{2}+\dfrac{7}{33}-\dfrac{18}{33}-\dfrac{5}{3}+\dfrac{48}{49}\)

\(=\dfrac{-9-10}{6}+\dfrac{-11}{33}+\dfrac{48}{49}\)

\(=\dfrac{-19}{6}+\dfrac{-1}{3}+\dfrac{48}{49}=-\dfrac{247}{98}\)

b: \(=\dfrac{11}{125}-\dfrac{17}{18}+\dfrac{4}{9}-\dfrac{5}{7}+\dfrac{17}{14}\)

\(=\dfrac{11}{125}+\dfrac{-17+8}{9}+\dfrac{-10+17}{14}\)

\(=\dfrac{11}{125}-1+\dfrac{7}{14}=\dfrac{11}{125}+\dfrac{1}{2}=\dfrac{22+125}{250}=\dfrac{147}{250}\)

\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)

=\(\frac{1}{2}\sqrt{3.4^2}-2\sqrt{3.5^2}-\sqrt{\frac{33}{11}}+5\sqrt{\frac{4}{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\sqrt{\frac{1}{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10}{3}\sqrt{3}\)

\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)

\(=\frac{-17}{3}\sqrt{3}\)

\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)

\(=\sqrt{6.5^2}+\sqrt{96}+4,5\sqrt{\frac{8}{3}}-\sqrt{6}\)

\(=5\sqrt{6}+\sqrt{6.4^2}+4,5\frac{\sqrt{24}}{3}-\sqrt{6}\)

\(=5\sqrt{6}+4\sqrt{6}+\frac{4,5.2\sqrt{6}}{3}-\sqrt{6}\)

\(=8\sqrt{6}+3\sqrt{6}\)

\(=11\sqrt{6}\)

Tự hòi tự trl :D ?

\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)

\(=\frac{1}{2}\sqrt{16.3}-2.5\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)

\(=-9\sqrt{3}+\frac{10}{3}\sqrt{3}=\left(-9+\frac{10}{3}\right)\sqrt{3}\)

\(=-\frac{17}{3}\sqrt{3}\)

\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\frac{2}{3}}-\sqrt{6}\)

\(=\sqrt{25.6}+\sqrt{1,6.60}+4,8\sqrt{\frac{8}{3}}-\sqrt{6}\)

\(=5\sqrt{6}+\sqrt{16.6}+4,5.\frac{1}{3}\sqrt{3^2.\frac{4.2}{3}}-\sqrt{6}\)

\(=9\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)

`=`\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{1}{3}-\dfrac{7}{8}\)

`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)

`= -21/24 = -7/8`

`b)`

\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)

`c)`

\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)

`d) `

\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)

`e)`

\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)

`= 17*(23-53+14-24)`

`= 17*(-40)`

`= -680`

`f)`

\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)

`= 19*(-218-82-533-167)`

`= 19*(-1000)`

`= -19000`

`g)`

\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)

`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)

`h)`

\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)

`i )`

\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)

`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)

`k)`

\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)

`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)

`l )`

\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)

P/s: Bạn tách bài ra hỏi nhé! Và ghi đề rõ ràng chứ đừng ghi ntnay, nhiều bạn nhìn vào rất khó nhìn!

`# \text {KaizulvG}`

a) 

`  1/2 - 5/6 + 11/33 - 35/40 `

`= 1/2 - 5/6 + 1/3 - 7/8 `

`= ( 1/2 - 5/6 ) + ( 1/3 - 7/8 ) `

`= ( 3/6 - 5/6 ) + ( 8/24 - 21/24 ) `

`= -2/6 + (-13)/24 `

`= -1/3 + (-13)/24 `

`= -8/24 + (-13)/24 `

`= -21/24 = -7/8 `

\(=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)

\(=\frac{1}{2}4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+\frac{10}{\sqrt{3}}=-9\sqrt{3}+\frac{10}{\sqrt{3}}=\frac{-17\sqrt{3}}{3}\)

a.\(\dfrac{27}{8}\)

b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)

d.\(\dfrac{7}{3}\)

e.5

g.\(\dfrac{53}{16}\)

Bài 1 :

a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)

b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)

c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)

d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)

g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)

a,3/2+5/4+5/8 = 12/8 + 10/8 + 5/8 = 27/8

b,4/5-3/8+2/4 = 32/40 - 15/40 + 20/40 = 37/40

c,3+6/8-5/4= 24/8 + 6/8 - 10/8 = 20/8 = 5/2

d,5/6-1/2+2 = 5/6 - 3/6 + 12/6 = 14/6 = 7/3

e,3/5+6/11+7/13+2/5+16/11+19/13 = ( 3/5+ 2/5) + ( 6/11 + 16/11) + ( 7/13 + 19/13) = 1 + 2 + 2 = 5

g,75/100+18/21+29/32+1/4+3/21+13/32 = 3/4 + 6/7 + 29/32 + 1/4 + 1/7 + 13/32=  ( 3/4 + 1/4) + ( 29/32 + 13/32) + ( 6/7 + 1/7)= 1 + 21/16 + 1 = 53/16

1/ \(7-2\sqrt{6}=\left(\sqrt{6}\right)^2-2\sqrt{6}+1\)

\(=\left(\sqrt{6}-1\right)^2\)

2/ \(10+2\sqrt{21}=\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2\)

\(=\left(\sqrt{7}+\sqrt{3}\right)^2\)

4/ \(10+4\sqrt{6}=2^2+2.2.\sqrt{6}+\left(\sqrt{6}\right)^2\)

\(=\left(2+\sqrt{6}\right)^2\)

5/ \(11-2\sqrt{30}=\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{5}+\left(\sqrt{5}\right)^2\)

= \(\left(\sqrt{6}-\sqrt{5}\right)^2\)

8/ \(11+4\sqrt{7}=2^2+2.2.\sqrt{7}+\left(\sqrt{7}\right)^2\)

= \(\left(2+\sqrt{7}\right)^2\)

10/ \(12+6\sqrt{3}=3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)

= \(\left(3+\sqrt{3}\right)^2\)

ks nhé

\(=\frac{1}{2}\sqrt{16.3}-2\sqrt{25.3}-\frac{\sqrt{3.11}}{\sqrt{11}}+5\sqrt{\frac{1.3+1}{3}}\)

\(=\frac{1}{2}\sqrt{4^2.3}-2\sqrt{5^2.3}-\frac{\sqrt{3}.\sqrt{11}}{\sqrt{11}}+5\sqrt{\frac{4}{3}}\)

\(=\frac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}}{\sqrt{3}}\)

\(=\frac{4}{2}\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}.\sqrt{4}}{\sqrt{3.}\sqrt{3}}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{2\sqrt{3}}{3}\)

\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\frac{\sqrt{3}}{3}\)

\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)

\(=-\frac{17}{3}\)