tính : \(\dfrac{\sqrt{48}+\sqrt{192}-\sqrt{175}}{\sqrt{3}}\)
Tính:
a. \(5\sqrt{2}-2\sqrt{48}+6\sqrt{75}-\sqrt{108}\)
b.\(2\sqrt{147}-\dfrac{3}{32}\sqrt{192}+\dfrac{4}{18}\sqrt{243}-\dfrac{1}{10}\sqrt{300}\)
c. \(-\dfrac{1}{2}\sqrt{108}+\dfrac{1}{15}\sqrt{75}-\dfrac{1}{22}\sqrt{363}+\sqrt{12}\)
d. \(\dfrac{5}{8}\sqrt{48}-\dfrac{1}{33}\sqrt{363}+\dfrac{3}{14}\sqrt{147}-\dfrac{1}{4}\sqrt{192}\)
e. \(\dfrac{3}{2}\sqrt{12}+\dfrac{7}{5}\sqrt{75}-\dfrac{9}{10}\sqrt{300}+\dfrac{11}{6}\sqrt{108}\)
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
Tính:
a) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
b) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
b: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8}{\sqrt{5}-1}\)
\(=2\sqrt{5}-2-2\sqrt{5}\)
=-2
c: \(=\dfrac{\sqrt{4}\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{\sqrt{6}}{2}\)
Rút gọn các biểu thức:
1. A= \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
2. B= \(\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\)\(\left(1-\sqrt{3}\right)\)
3. C= (\(2\sqrt{7}-2\sqrt{6}\) ) . \(\sqrt{6}\) - \(\sqrt{168}\)
4. D=( \(\sqrt{28}-2\sqrt{8}+\sqrt{7}\) ). \(\sqrt{7}+4\sqrt{14}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
4) A = 3.\(\sqrt{2}\)+ 4.\(\sqrt{8}\)-\(\sqrt{18}\)
5) ( \(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\) ) :2\(\sqrt{3}\)
4. \(A=3\sqrt{2}+4\sqrt{8}-\sqrt{18}=3\sqrt{2}+8\sqrt{2}=3\sqrt{2}=8\sqrt{2}\)
5. \(\dfrac{\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}}{2\sqrt{3}}=\dfrac{2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}}{2\sqrt{3}}=\dfrac{-16\sqrt{3}}{2\sqrt{3}}=-8\sqrt{3}\)
4: Ta có: \(A=3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}\)
\(=8\sqrt{2}\)
5: Ta có: \(\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\)
\(=\left(2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):2\sqrt{3}\)
\(=-16\sqrt{3}:2\sqrt{3}=-8\)
a. \(\dfrac{1}{3}.\sqrt{18}-\sqrt{192}-\dfrac{\sqrt{33}}{\sqrt{11}}+3\sqrt{5\dfrac{1}{3}}\)
b. \(\sqrt{\left(2\sqrt{3}-5\right)^2}-2\sqrt{7+4\sqrt{3}}\)
a: \(\dfrac{1}{3}\cdot\sqrt{18}-\sqrt{192}-\dfrac{\sqrt{33}}{\sqrt{11}}+3\cdot\sqrt{5\dfrac{1}{3}}\)
\(=\dfrac{1}{3}\cdot3\sqrt{2}-8\sqrt{3}-\sqrt{3}+3\cdot\dfrac{4}{\sqrt{3}}\)
\(=\sqrt{2}-7\sqrt{3}+4\sqrt{3}\)
\(=\sqrt{2}+3\sqrt{3}\)
b: Ta có: \(\sqrt{\left(2\sqrt{3}-5\right)^2}-2\cdot\sqrt{7+4\sqrt{3}}\)
\(=5-2\sqrt{3}-2\cdot\left(2+\sqrt{3}\right)\)
\(=5-2\sqrt{3}-4-2\sqrt{3}\)
\(=-4\sqrt{3}+1\)
B1. ko sử dụng máy tính, rút gọn
\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
B2.
\(G=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
so sánh G với 1
B3. giải pt
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
Bài 1:
\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)
\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)
\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}\right)^2}-\sqrt{2}=\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}+\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)
Bài 2:
Ta có: G-1
\(=\dfrac{\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\dfrac{-\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ
hay \(G\le1\)
Thực hiện phép tính và thu gọn biểu thức:
B= \(\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
Thực hiện phép tính:
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
Tính
a) \(2\sqrt[3]{24}-5\sqrt[3]{81}+4\sqrt[3]{192}\)
b) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
c) \(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{3}-\dfrac{1}{\sqrt[3]{2}+1}\)
b: Ta có: \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}\cdot\sqrt[3]{4}\)
\(=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54\cdot4}\)
=3-6
=-3
1)\(\sqrt{12}\)\(-\)\(\sqrt{27}\)\(+\)\(\sqrt{48}\)
2)(\(\sqrt{24}+\sqrt{20}-\sqrt{80}\))\(\div\)5
3)2\(\sqrt{27}-\sqrt{\dfrac{16}{3}}\)\(-\)\(\sqrt{48}-\)\(\sqrt{8\dfrac{1}{3}}\)
4) \(\dfrac{1}{\sqrt{5}-\sqrt{3}}\)\(-\)\(\dfrac{1}{\sqrt{5+\sqrt{3}}}\)
\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
1) \(=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\)
2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)
3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
4) \(=\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{5-3}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)