2.(−7)2+3.(−4)3−60
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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) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
A=(2+2^2)+...+(2^59+2^60)
=2(1+2)+...+2^59(1+2)
=3(2+2^3+...+2^59)
nên A chia hết cho 3.
A= (2+2^2+2^3)+...+(2^58+2^59+2^60)
=2(1+2+2^2)+...+2^58(1+2+2^2)
=7(2+2^4+..+2^58)
nên A chia hết cho 7
A= (2+2^2+2^3+2^4)+....+(2^57+2^58+2^59+2^6...
=2(1+2+2^2+2^3)+....+2^57(1+2+2^2+2^3)...
=15(2+2^5+...+2^57)
nên A chia hết cho 15
2S=32+33+34+....+32016
2S-S=(32+33+34+...+32016)-(3+32+33+....+32015)
S=22016-3
4(x-3)2-320 = 0
=> 4(x-3)2 = 320
=> (x-3)2 = 320 : 4 = 80 = (8,94427191)2 = (-8,94427191)2
TH1:
x - 3 = 8,94427191
=> x = 11,94427191
TH2:
x - 3 = -8,94427191
=> x = -5,94427191
7(4+x)3-875 = 0
=> 7(4+x)3 = 875
=> (4+x)3 = 875:7 = 125 = 53
=> 4 + x = 5
=> x = 1
650 - 5(x+4)2 = 330
5(x+4)2 = 650 - 330 =320
=> (x+4)2 = 320 : 5 = 64 = 82
=> x+4 = 8
=> x = 4
3(5-x)2-15 = 60
=> 3(5-x)2 = 75
=> (5-x)2 = 25 = 52 =(-5)2
TH1:
5-x =5
=> x = 0
TH2: 5-x = -5
=> x = 10
`2.(−7).2+3.(−4).3−60`
`=4.(-7)+9.(-4)-60`
`=(-28)+(-36)-60`
`=-124`