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16 tháng 6 2023

\(B=50-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)

\(=50-3.\sqrt{7^2.2}+2\sqrt{2^2.2}+3\sqrt{4^2.2}-5\sqrt{3^2.2}\)

\(=50-3.7\sqrt{2}+2.2\sqrt{2}+3.4\sqrt{2}-5.3\sqrt{2}\)

\(=50-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{2}\)

\(=50+\sqrt{2}.\left(-21+4+12-15\right)\)

\(=50+\sqrt{2}.\left(-20\right)\)

\(=50-20\sqrt{2}\)

\(C=\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{5}-\sqrt{7}\right)\)

\(=\left(\sqrt{3}+\sqrt{5}\right)^2-\sqrt{7}^2\)

\(=\sqrt{3}^2+2.\sqrt{3}.\sqrt{5}+\sqrt{5}^2-7\)

\(=2\sqrt{15}+3+5-7\)

\(=2\sqrt{15}+1\)

Nghĩ ra xong tính thử thấy đúng định nàm xong thấy mẹ giải r ấy:")). Với nại con còn nhỏ nắm, hong bic nhiều cái mà nớp 9 hay sử dụng nữa ý, sợ dùng sai;-;.

23 tháng 8 2023

a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)

\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)

\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)

\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)

\(=-8\sqrt{3}\)

b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)

\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)

30 tháng 5 2022

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

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

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

15 tháng 11 2022

a: \(=9\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}=18\sqrt{2}\)

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

c: \(=2\sqrt{21}\)

 

6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)

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

8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)

9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

21 tháng 9 2020

Bài 2 : 

a) \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+2\sqrt{7}+1}-\sqrt{7}\)

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

b) \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{4+4\sqrt{3}+3}-2\sqrt{3}\)

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

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

c) \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)

\(=\sqrt{13-2\sqrt{13}+1}+\sqrt{13+2\sqrt{13}+1}\)

\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)

\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)

\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)

d) \(D=\sqrt{22-2\sqrt{21}}+\sqrt{22+2\sqrt{21}}\)

\(=\sqrt{21-2\sqrt{21}+1}+\sqrt{21+2\sqrt{21}+1}\)

\(=\sqrt{\left(\sqrt{21}-1\right)^2}+\sqrt{\left(\sqrt{21}+1\right)^2}\)

\(=\left|\sqrt{21}-1\right|+\left|\sqrt{21}+1\right|\)

\(=\sqrt{21}-1+\sqrt{21}+1=2\sqrt{21}\)

21 tháng 9 2020

bạn j ơi bạn giải đúng k vậy

(10−15+33)5−72

=15−15+15−62

 

(1550+5200−3450)810

=(15.52+5.102−3.152)810