a)205-510/1005 b)46x95+69x120/84x312-611
Những câu hỏi liên quan
thực hiện phép tính
(0,125)3✖83 +(3/4 -5/6 )2 -503\125 + 205✖510\1005
thực hiện phép tính
(0,125)3✖83 +(3/4 -5/6 )2 -503\125 + 205✖510\1005
thực hiện phép tính
(0,125)3✖83 +(3/4 -5/6 )2 -503\125 + 205✖510\1005
thực hiện phép tính
(0,125)3✖83 +(3/4 -5/6 )2 -503\125 + 205✖510\1005
cho a/b=c/d.cmr:a^1005+b^1005/c^1005+d^1005=(a+b)^1005/(c+d)^1005
ta có \(\frac{a}{b}=\frac{c}{d}\)
=>\(\frac{a}{c}=\frac{b}{d}\)(1)
Từ (1) => \(\frac{a^{1005}}{c^{1005}}=\frac{b^{1005}}{d^{1005}}=\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}\)(2)
Từ (1) => \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
=>\(\left(\frac{a}{c}\right)^{1005}=\left(\frac{b}{d}\right)^{1005}=\left(\frac{a+b}{c+d}\right)^{1005}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)(3)
mà \(\left(\frac{a}{c}\right)^{1005}=\frac{a^{1005}}{c^{1005}}\)
từ 2 zà 3 => ghi lại cái cần chứng minh nha ( dpcm)
Cho a , b ,c thỏa mãn a^2010 + b^2010 + x^2010 = a^1005.b^1005 + b^1005.c^1005 + c^1005 a^1005 Tính (a - b)^20 + (b - c)^11 + (c - a)^2010
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Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
Vậy (a - b)20 + (b - c)11 + (c - a)2010 = (a - a)20 + (a - a)11 + (a - a)2010 = 0 + 0 + 0 = 0 .
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Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
Vậy (a - b)20 + (b - c)11 + (c - a)2010
= (a - a)20 + (a - a)11 + (a - a)2010
= 0 + 0 + 0
= 0 .
=> ĐPCM
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Xem thêm câu trả lời
Cho a , b ,c thỏa mãn a^2010 + b^2010 + c^2010 = a^1005.b^1005 + b^1005.c^1005 + c^1005 a^1005 Tính (a - b)^20 + (b - c)^11 + (c - a)^2010
Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
Vậy (a - b)20 + (b - c)11 + (c - a)2010 = (a - a)20 + (a - a)11 + (a - a)2010 = 0 + 0 + 0 = 0 .
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a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
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Chứng minh rằng : \(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
Cho a, b, c là các số thực. Chứng minh:
a2010 + b2010 + c2010> a1005b1005 + a1005c1005 + c1005b1005
Ta có \(\left(a^{1005}-b^{1005}\right)^2+\left(b^{1005}-c^{1005}\right)^2+\left(c^{1005}-a^{1005}\right)^2>0\Leftrightarrow a^{2010}-2a^{1005}b^{1005}+b^{2010}+b^{2010}-2b^{1005}c^{1005}+c^{2010}+c^{2010}-2a^{1005}c^{1005}+a^{1005}>0\Leftrightarrow2\left(a^{2010}+b^{2010}+c^{2010}\right)-2\left(a^{1005}b^{1005}+a^{1005}c^{1005}+c^{1005}b^{1005}\right)>0\Leftrightarrow a^{2010}+b^{2010}+c^{2010}>a^{1005}b^{1005}+a^{1005}c^{1005}+c^{1005}b^{1005}\)(đpcm)
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cho tỉ lệ thức a/b = c/d. cmr ta có tỉ lệ thức sau: \(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)