Tính nhanh:
\(A=\frac{2010X2011-1005}{2010X2010+1005}\)
Những câu hỏi liên quan
Tính nhanh:
\(A=\frac{1006\times2013-1005}{1005\times2013+1006}\)
=1 muốn biết cách làm thì kết bạn vs mình nha
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Tính nhanh:
\(A=\frac{1006\times2013-1005}{1005\times2013+1006}\)
A = \(\frac{1006.2013-1005}{1005.2013+1006}\)
A = \(\frac{\left(1005+1\right).2013-1005}{1005.2013+1006}\)
A = \(\frac{1005.2013+2013-1005}{1005.2013+1006}\)
A = \(\frac{1005.2013+1006}{1005.2013+1006}\)
A = 1
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Chứng minh rằng : \(\frac{a^{1005}+b^{^{1005}}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
Giúp mình nhanh nha , cần gấp, đúng mik tích cho
ta có a^1005+b^1005 / c^1005+d^1005
=> a^1005/c^1005=b^1005/d^1005
=a/c=b/d=a+b/c+d=(a+b)^2015/(c+d)^1005
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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 tỉ lệ thức a/b = c/d. cmr ta có tỉ lệ thức sau: \(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\) chứng minh rằng
a) (a+2c).(b+d)=(a+c).(b+2d)
b)\(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}\)=\(\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng:
a.\(\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\) b.\(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\\ =>\orbr{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(Taco:\left(a+2c\right).\left(b+d\right)=\left(a+c\right).\left(b+2d\right)\)
\(=>\left(bk+2dk\right).\left(b+d\right)=\left(bk+dk\right).\left(b+2d\right)\)
\(=>\frac{bk+2dk}{bk+dk}=\frac{b+2d}{b+d}\)
\(=>\frac{k.\left(b+2d\right)}{k.\left(b+d\right)}=\frac{b+2d}{b+d}\)
\(=>\frac{b+2d}{b+d}=\frac{b+2d}{b+d}\)(ĐPCM)
, Chờ tí mk làm câu b
Ta có :\(\frac{a}{b}=\frac{c}{d}\)
\(\implies\)\(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}\left(1\right)\) \(\implies\) \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\left(2\right)\)
Từ (1);(2)\(\implies\) \(\frac{a+2c}{b+2d}=\frac{a+c}{b+d}\)
\(\implies\) \(\left(a+2c\right).\left(b+d\right)=\left(b+2d\right).\left(a+c\right)\)
P/S : ko chắc
Áp dụng tc của dãy tỉ số bằng nhau có :
\(\frac{a}{b}=\frac{c}{d}=\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)(ĐPCM)
Đánh máy ẩu v :D
Xem thêm câu trả lời
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
https://olm.vn/hoi-dap/question/1038454.html
Mình vừa làm cách đây 11 phút nhé !
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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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