Rút gọn biểu thức
a) \(2\sqrt{28}+\sqrt{63}-\sqrt{112}\)
b) \(3\sqrt{48}-5\sqrt{108}+6\sqrt{\dfrac{1}{3}}\)
Rút gọn các biểu thức sau:
a) \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
b) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{10}{1+\sqrt{6}}\)
a: Ta có: \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}-2\sqrt{5}\)
\(=-\sqrt{7}-2\sqrt{5}\)
rút gọn biểu thức sau
a,\(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
b,\(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)
\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)
\(=14\sqrt{3}-2\sqrt{57}\)
b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)
\(=\left(4-6+3-5\right)\sqrt{6}\)
\(=-4\sqrt{6}\)
Rút gọn các biểu thức sau:
a) $E=2 \sqrt{40 \sqrt{12}}+3 \sqrt{5 \sqrt{48}}-2 \sqrt{\sqrt{75}}-4 \sqrt{15 \sqrt{27}}$ :
b) $F=\dfrac{1}{\sqrt{3}}+\dfrac{1}{3 \sqrt{2}}+\dfrac{1}{\sqrt{3}} \sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}} .$
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
Rút gọn các biểu thức sau:
a) \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}\)
b) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
c) \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}\)
d) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e) \(\dfrac{\sqrt{a}-\sqrt{b}^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\) với a > 0, b > 0
a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)
c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)
d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)
\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)
e, Ghi đúng đề.
\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
Bài 1.Rút gọn A = \(\sqrt{x^2+\dfrac{2x^2}{3}}\) với x<0
Bài 2.Rút gọn biểu thức \(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right)\):\(\dfrac{2}{2\sqrt{5}-\sqrt{6}}\)
Bài 3.Cho ba biểu thức A = a\(\sqrt{b}\) + b\(\sqrt{a}\);B = \(a\sqrt{a}-b\sqrt{b}\) ;C = a-b.Trong ba biểu thức trên biểu thức bằng biểu thức \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\) với a,b>0
Bài 7.Cho B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{100}}\).Giá trị của biểu thức B là
Bài 8.Gọi M là giá trị nhỏ nhất của \(\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\) và N là giá trị lớn nhất của \(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\).Tìm M và N
Giúp mình với!Mình đang cần gấp
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
Rút gọn các biểu thức sau :
a) A= \(\sqrt{18}\) . \(\sqrt{2}\) - \(\sqrt{48}\) : \(\sqrt{3}\)
b)B= \(\dfrac{8}{\sqrt{5}-1}\) + \(\dfrac{8}{\sqrt{5}+1}\)
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
Rút Gọn
1.\(3\sqrt{2}-4\sqrt{18}-2\sqrt{32}-\sqrt{50}\)
2,\(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
3,\(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
4,\(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
5,\(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}\)
6,\(10\sqrt{28}-2\sqrt{275}-3\sqrt{343}-\frac{3}{2}\sqrt{396}\)
tính
1.\(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
2.\(\sqrt{28}-4\sqrt{63}+7\sqrt{112}\)
3.\(\sqrt{49}-5\sqrt{28}+\dfrac{1}{2}\sqrt{63}\)
4.\(\left(2\sqrt{6}-4\sqrt{3}-\dfrac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
5.(\(2\sqrt{1\dfrac{9}{16}}-5\sqrt{5\dfrac{1}{16}}\)):\(\sqrt{16}\)
6.\(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt{3}\)
7.\(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt{2}-\sqrt{162}:\sqrt{2}\)
8.\(\left(2\sqrt{1\dfrac{9}{10}}-\sqrt{5\dfrac{1}{10}}\right):\sqrt{10}\)
9.\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
10.\(2\sqrt{27}-6\sqrt{\dfrac{4}{3}}+\dfrac{3}{5}\sqrt{75}\)
11.\(\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}\)
12.\(\dfrac{\sqrt{27}}{\sqrt{3}}+\dfrac{\sqrt{98}}{\sqrt{2}}-\sqrt{175}:\sqrt{7}\)
13.\(\left(\dfrac{\sqrt{8}}{\sqrt{2}}-\dfrac{\sqrt{180}}{\sqrt{5}}\right).\sqrt{5}-\sqrt{\dfrac{81}{11}}.\sqrt{11}\)
14.\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
15.\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
16.\(\left(1+\sqrt{5}-\sqrt{3}\right)\left(1+\sqrt{5}+\sqrt{3}\right)\)
Câu 1: Rút gọn các biểu thức sau:
A= (\(\sqrt{48}\) - 2\(\sqrt{3}\) + 2\(\sqrt{5}\) )\(\sqrt{5}\) - 2\(\sqrt{45}\)-\(\sqrt{3}\)
B= (\(\dfrac{1}{\sqrt{5}-\sqrt{2}}\)\(-\dfrac{1}{\sqrt{5}+\sqrt{2}}\)).\(\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
C= \(2\sqrt{a}\)-\(\sqrt{9a^3}\) a\(^2\) \(\sqrt{\dfrac{4}{a}}\)+ \(\dfrac{2}{a^2}\)\(\sqrt{25a^3}\) với a > 0
a: Ta có: \(A=\left(\sqrt{48}-2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-2\sqrt{45}-\sqrt{3}\)
\(=\left(2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-6\sqrt{5}-\sqrt{3}\)
\(=2\sqrt{15}+10-6\sqrt{5}-\sqrt{3}\)
b: Ta có: \(B=\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}-\dfrac{1}{\sqrt{5}+\sqrt{2}}\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{2}}{9+6\sqrt{2}}=\dfrac{-8+6\sqrt{2}}{3}\)