1a) 6\(\sqrt{3}\) - 5\(\sqrt{12}\) + 3\(\sqrt{75}\)
b) 2\(\sqrt{5}\) - \(\dfrac{1}{4}\) \(\sqrt{80}\) + 7\(\sqrt{500}\)
c) \(\dfrac{sin43^o}{cos47^o}\) + tan45o
d) \(\dfrac{tan32^o}{tan68^o}\) - cos30o - \(\dfrac{sin18^o}{sin82^o}\)
Tính giá trị biểu thức
a,\(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)
b,\(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)
c,\(\tan^2\)\(40^o\)*\(sin^250^o-3+\left(1-sin40^o\right)\left(1+sin40^o\right)\)
a: \(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)
\(=6\sqrt{5}+\sqrt{5}-12\sqrt{5}\)
\(=-5\sqrt{5}\)
b: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\)
\(=-8\sqrt{3}+1\)
Bài 1 :
a, \(\dfrac{1}{2}\sqrt{12}+\sqrt{27}-\sqrt{75}\)
b, \(\sqrt{7-4\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
c, 6\(\sqrt{27}-2\sqrt{75}-\dfrac{1}{2}\sqrt{300}\)
d, \(\dfrac{7}{\sqrt{10}-\sqrt{3}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}-\dfrac{6}{\sqrt{3}}\)
e, \(\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.(3\sqrt{2}+\sqrt{14)}\)
f, \(\sqrt{11-4\sqrt{ }7}+\dfrac{2\sqrt{7}-2}{\sqrt{7}-1}\)
g, \((\sqrt{125}-3\sqrt{3})\dfrac{\sqrt{5}-\sqrt{3}}{8+\sqrt{15}}\)
h, \(\sqrt{100}-\sqrt{64}\)
i, \(\sqrt{(1-\sqrt{3})^2}-\sqrt{3}\)
Bạn nào biết làm bài này thì giúp mình với ạ ! sáng mai mình cần gấp !
a: \(=\dfrac{1}{2}\cdot2\sqrt{3}+3\sqrt{3}-5\sqrt{3}=-\sqrt{3}\)
b: \(=2-\sqrt{3}-\sqrt{3}-1=1\)
c: \(=18\sqrt{3}-10\sqrt{3}-\dfrac{1}{2}\cdot10\sqrt{3}=3\sqrt{3}\)
d: \(=\sqrt{10}+\sqrt{3}-\sqrt{5}+\sqrt{2}-2\sqrt{3}=\sqrt{10}+\sqrt{2}-\sqrt{3}-\sqrt{5}\)
RÚt gọn c)\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{899}+\sqrt{900}}\)
d)\(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
e) \(\dfrac{x+5-x\sqrt{x-1}}{x-1-3\sqrt{x-1}}\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}\)
=)) giúp câu nào được thì giúp ạ <3
d) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\dfrac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}+\dfrac{6\left(3-\sqrt{3}\right)}{6}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}=3\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}=\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{20+2\sqrt{20}+1}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{20}+1\right)^2}=\sqrt{5}-\left(\sqrt{20}+1\right)=\sqrt{5}-2\sqrt{5}-1=-1-\sqrt{5}\)
c) \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{899}+\sqrt{900}}\)
\(=\dfrac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}+\dfrac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-\sqrt{4}\right)}+...+\dfrac{\sqrt{899}-\sqrt{900}}{\left(\sqrt{899}+\sqrt{900}\right)\left(\sqrt{899}-\sqrt{900}\right)}\)\(=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+...+\dfrac{\sqrt{899}-\sqrt{900}}{899-900}\)
\(=-\left(1-\sqrt{2}\right)-\left(\sqrt{2}-\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{4}\right)+...-\left(\sqrt{899}-\sqrt{900}\right)\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+...-\sqrt{899}+\sqrt{900}\)
\(=-1+\sqrt{900}\) \(=-1+30=29\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{5}-\sqrt{\sqrt{20}+2\sqrt{20}+1}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{20}+1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{20}+1\right)\)
\(=\sqrt{5}-\sqrt{20}-1\)
\(=\sqrt{5}-2\sqrt{5}-1\)
\(=-\sqrt{5}-1\)
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:
\(a.\) \(A=\sqrt{12}-2\sqrt{48}+\dfrac{7}{5}\sqrt{75}\)
\(b.\) \(B=\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(c.\) \(C=\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
\(d.\) \(D=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}-5}{\sqrt{5}}-\dfrac{11}{2\sqrt{5}+3}\)
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
tính
A=\(\left(1-\sqrt{7}\right).\dfrac{\sqrt{7}+7}{2\sqrt{7}}\)
B=\(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
C=\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
D=\(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)
E=\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
a \(\left(2\sqrt{6-}4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)5\sqrt{6}\)
b \(\left(\dfrac{1}{\sqrt{5-}\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}+1\right).\dfrac{1}{\sqrt{2}+1^2}\)
c \(\dfrac{3}{\sqrt{2+}\sqrt{7}}\dfrac{1}{\sqrt{2-}\sqrt{7}}-1\)
d \(2x\sqrt{300}-15\sqrt{75}+5\sqrt{75}.15\)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
Bài 1 Rút gọn
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) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
Bài 2: Cho 2 đường thẳng (d): y = -x - 4 và (d₁): y = 3x + 2.
a) Vẽ đồ thị (d) và (d₁) trên cùng một mặt phẳng tọa độ Oxy.
b) Xác định tọa điểm A của 2 đường thẳng trên.
c) Viết pt đường thẳng: (d₂): y = ax + b (a≠0) song song vs đường thẳng (d) và đi qua điểm B(-2;5)
Câu 1:
a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)
\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)
\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)
\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)
\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Bài 2:
a:
b: Phương trình hoành độ giao điểm là:
\(3x+2=-x-4\)
=>4x=-6
=>x=-3/2
Thay x=-3/2 vào y=-x-4, ta được:
\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)
Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)
c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)
Vậy: (d2): y=-x+b
Thay x=-2 và y=5 vào (d2), ta được:
\(b-\left(-2\right)=5\)
=>b+2=5
=>b=5-2=3
Vậy: (d2): y=-x+3
Tính:
a) \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
b) \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)
c) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)