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a. \(\dfrac{\sqrt{444}}{\sqrt{111}}=\sqrt{\dfrac{444}{111}}=\sqrt{4}=2\)
b. \(\sqrt{75}-\sqrt{27}-\sqrt{108}\)
\(=5\sqrt{3}-3\sqrt{3}-6\sqrt{3}\)
\(=-4\sqrt{3}\)
\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)
\(=9\sqrt{5}-28\sqrt{3}\)
\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(=4-2\cdot5+6-7\)
\(=4-10+6-7\)
=-7
A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)
=\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)
=2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)
=\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)
=9\(\sqrt{5}\)- 28\(\sqrt{3}\)
\(999+888+777+666+555+444+333+222+111\)
\(=\left(999+111\right)+\left(888+222\right)+\left(777+333\right)+\left(666+444\right)+555\)
\(=1110+1110+1110+1110+555\)
\(=\left(1110\times4\right)+555\)
\(=4440+555\)
\(=4995\)
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)
a: \(=\dfrac{1}{2}\cdot4\sqrt{3}+4\cdot3\sqrt{3}-2\cdot6\sqrt{3}\)
\(=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=2\sqrt{3}\)
b: \(=\left|2-\sqrt{5}\right|-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(=\sqrt{5}-2-3-\sqrt{5}\)
=-5
Đề sai: Ví dụ m = 1 => B = \(\sqrt{46}\) không là số nguyên
Sửa đề: B = \(\sqrt{444...4+444...4+1}\)
B2 = 444....4 + 444....4 + 1
Đặt k = 111...1 (m chữ số 1 ) => 9k = 999..9 (m chữ số 9 ) = 10m - 1 => 10m = 9k + 1
Ta có : 999...9 (2m chữ số 9 ) = 9 x 111....1 (2m chữ số ) = 102m - 1
=> 111..1 (2m chữ số 1) = \(\frac{10^{2m}-1}{9}\)=> 444...4 (2m chữ số 4 ) = \(\frac{4.\left(10^{2m}-1\right)}{9}=\frac{4.\left(\left(9k+1\right)^2-1\right)}{9}=\frac{4}{9}.\left(81k^2+18k\right)=36k^2+8k\)
Ta có: B2 = 36k2 + 8k + 4.k + 1 = 36k2 + 12 k + 1 = (6k + 1)2 => B = 6k + 1 là số nguyên => đpcm
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: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
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) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)
\(=\left(4-6+3-5\right)\sqrt{6}\)
\(=-4\sqrt{6}\)
a \(\dfrac{\sqrt{444}}{\sqrt{111}}=\dfrac{\sqrt{4\times111}}{\sqrt{111}}=\dfrac{2\sqrt{111}}{\sqrt{111}}=2\)
b\(\sqrt{75}-\sqrt{27}-\sqrt{108}=\sqrt{25\times3}-\sqrt{9\times3}-\sqrt{36\times3}=5\sqrt{3}-3\sqrt{3}-6\sqrt{3}=-4\sqrt{3}\)