Câu 1: Tính
a. \(\sqrt{3\dfrac{6}{25}}\) b. \(\sqrt[3]{261}\) c. \(\sqrt{8,1}\) . \(\sqrt{20}\). \(\sqrt{8}\)
d. \(\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}\)
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
* Tính
a. A=\(\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
b. B=\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1
thực hiện phép tính
a, \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
b, \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
c, \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).
\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=\dfrac{\left(\sqrt{2-\sqrt{3}}\right)^2+\left(\sqrt{2+\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}=\dfrac{4}{1}=4\)
tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)
6) Thực hiện các phép tính
a. \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
b. \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
c. \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
Mng giúp mik vs ạ >.<!!
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)
\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)
c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
\(=\left(2+\sqrt{3}\right)\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)
\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)
1) thực hiện phép tính
a) \(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
b) \(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
c) \(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}\)
giúp mk vs ạ mk đang cần gấp
a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)
\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)
c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)
1. Tính
a) \(\sqrt[3]{(\sqrt{2}+3)(11+6\sqrt{2})}\sqrt[3]{(\sqrt{2}+-3)(11-6\sqrt{2})}\)
b) (\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}-\sqrt[3]{2})\)
c)\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
Bài 1: Tính
a) \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
b) \(1\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
c) \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
Bài 2: Cho (d₁): y = \(\dfrac{1}{2}x-4\) và (d₂): y = \(-3x+3\) . Vẽ (d₁) và (d₂) trên cùng một hệ trục tọa độ. Tìm tọa độ giao điểm A của 2 đường thẳng trên.
Helpp!!
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
* Thực hiện phép tính
a. \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{2\left(-5\right)^2}\)
b. \(\dfrac{\sqrt{270}}{\sqrt{30}}-\sqrt{1,8}.\sqrt{20}\)