135,67 - 65,34 - 35,67= 34,66

^_^ k nhaNguyenx Văn Tâm

135,67 - 65,34 - 35,67=

(135,67 - 25,67) - 65,34=

100 - 65,34=

34,66

= 59,07

59,07

\(x+35,67=88,5\)

\(x=88,5-35,67\)

\(x=52,83\)

Đáp án B 

Ta có:  35 , 67 < x < 36 , 05  và x là số tự nhiên nên x = 36

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

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

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

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

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=\sqrt{7}+\sqrt{7}=2\sqrt{7}\)

Ta có: \(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

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

\(=\sqrt{\sqrt{7}^2-2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}+\sqrt{\sqrt{7}^2+2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}\)

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

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

4/7 *1/3+5/7:2/3

\(\frac{4}{21}\)+ \(\frac{15}{14}\)= \(\frac{53}{42}\)

=4/21+15/14

=53/42

a, =20/5=4

b,=247+47=294

tick nha

a) Ta có: \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}-\dfrac{20}{\sqrt{10}}\)

\(=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}-\dfrac{6\left(\sqrt{10}+2\right)}{\left(\sqrt{10}+2\right)\cdot\left(\sqrt{10}-2\right)}-\dfrac{20}{\sqrt{10}}\)

\(=\sqrt{10}-6\sqrt{10}-12-2\sqrt{10}\)

\(=-7\sqrt{10}-12\)

b) Ta có: \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

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

\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)\)

=5-9=-4

c) Ta có: \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)

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

\(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)

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

\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)

d) Ta có: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)

\(=4+\sqrt{11}+\dfrac{3-\sqrt{7}}{2}-2\sqrt{7}-4-\dfrac{\sqrt{7}-5}{2}\)

\(=\sqrt{11}-2\sqrt{7}+\dfrac{3-\sqrt{7}-\sqrt{7}+5}{2}\)

\(=\dfrac{2\sqrt{11}-4\sqrt{7}+8-2\sqrt{7}}{2}\)

\(=\dfrac{2\sqrt{11}-6\sqrt{7}+8}{2}\)

\(=\sqrt{11}-3\sqrt{7}+4\)

e) Ta có: \(\dfrac{2\sqrt{12}-\sqrt{6}}{2\sqrt{6}-\sqrt{3}}+\dfrac{10+\sqrt{5}}{2\sqrt{15}+\sqrt{3}}\)

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

\(=\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{2}+\sqrt{15}}{3}\)