Cho A = 3^60 - 3^59 +3^58 -3^57 + .....+3^2 - 3
Chúng tỏ : A=3^61 - 3 : 4
A=360-359+358-357+.....+32-3
Chứng tỏ A=361-3/4
\(A=\left(3^{60}+3^{58}+3^{56}+...+3^2\right)-\left(3^{59}+3^{57}+3^{55}+...+3\right).\)
\(B=3^{60}+3^{58}+3^{56}+...+3^2\)
\(9B=3^{62}+3^{60}+3^{58}+...+3^4\)
\(B=\frac{9B-B}{8}=\frac{3^{62}-3^2}{8}=\frac{3^2\left(3^{60}-1\right)}{8}\)
\(C=3^{59}+3^{57}+3^{55}+...+3\)
\(9C=3^{61}+3^{59}+3^{57}+...+3^3\)
\(C=\frac{9C-C}{8}=\frac{3^{61}-3}{8}=\frac{3\left(3^{60}-1\right)}{8}\)
\(A=B-C=\frac{3^2\left(3^{60}-1\right)-3\left(3^{60}-1\right)}{8}=\frac{6\left(3^{60}-1\right)}{8}\)
\(A=\frac{2.3.\left(3^{60}-1\right)}{8}=\frac{2.3.3^{60}}{8}-\frac{2.3}{8}=\frac{3^{61}}{4}-\frac{3}{4}=\frac{3^{61}-3}{4}\)
a) 57+58+59+60+61-17-18-19-20-21
b) 9-10+11-12+13-14+15-16
c) -1-2-3-4-5-.....-1999-2000-2001-2002
F = 1/59 + 2/58 + 3/57 + ... + 58/2 + 59/1 : 1/2 + 1/3 + ... + 1/60
Chứng tỏ rằng giá trị của tổng sau luôn chia hết cho 31
A = 5 +5^2+5^3+5^4+5^4+5^6+....+5^58+5^59+5^60
a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+3+x+4=0\)
\(\Leftrightarrow-x+7=0\)
\(\Leftrightarrow-x=-7\)
hay x=7
Vậy: S={7}
b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow-6x+8=-10x+10\)
\(\Leftrightarrow-6x+8+10x-10=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
nên x-60=0
hay x=60
Vậy: S={60}
a) 371 + 731 - 271 -531
b) 57 + 58 + 59 + 60 + 61 - 17 - 18 -19 -20 -21
c) 9 - 10 + 11 - 12 + 13 - 14 + 15 - 16
d) - 1 - 2 - 3 - ..... - 2005 - 2006 - 2007
a) tự làm
b) (57-17)+(58-18)+(59-19)+(60-20)+(61-21)=40+40+40+40+40=200
c) (9+15)+(11+13)+(-10-14)+(-12-12-4)=24+24-24-24-4=-4
d)-(1+2+3+.....+2007)=\(-\dfrac{\left(1+2007\right).2007}{2}=-2015028\)
50. √(98-16√3)
51. √(2-√3)
52. √(4+√15)
53. √(5-√21)
54. √(6-√35)
55. √(2+√3)
56. √(4-√15)
57. √(8-√55)
58. √(7+√33)
59. √(6+√35)
60. √(7-3√5)
50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)
51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)
54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)
55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
Tính:
M=60-59+58-57+56-55+........+4-3+2-1
M = 60 - 59 + 58 - 57 + 56 - 55 + ... + 4 - 3 + 2 - 1
M = ( 60 - 59 ) + ( 58 - 57 ) + ( 56 - 55 ) + ... + ( 4 - 3 ) + ( 2 - 1 )
M = 1 + 1 + 1 + ... + 1 + 1
Vì tổng M có 60 số hạng,mà 2 số hạng tạo thành 1 cặp nên 60 số hạng tạo thành 30 cặp
M = 1 . 30
M = 30
tính tổng:
\(\frac{\sqrt{3}}{\sqrt[3]{2}+\sqrt[3]{4}}+\frac{\sqrt{5}}{\sqrt[3]{4}+\sqrt[3]{6}}+\frac{\sqrt{7}}{\sqrt[3]{6}+\sqrt[3]{8}}+...+\frac{\sqrt{57}}{\sqrt[3]{56}+\sqrt[3]{58}}+\frac{\sqrt{59}}{\sqrt[3]{58}+\sqrt[3]{60}}\)