cho tỉ lệ thức a/b = c/d. cmr ta có tỉ lệ thức sau: \(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
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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
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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 \(\frac{a}{b}\)=\(\frac{c}{d}\) chứng minh rằng
a) (a+2c).(b+d)=(a+c).(b+2d)
b)\(\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}\)=\(\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)
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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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 giá trị biểu thức \(M=\left(a-b\right)^{20}+\left(b-c\right)^{12}+\left(c-a\right)^{2013}\)
Đặt \(\left\{{}\begin{matrix}a^{1005}=x\\b^{1005}=y\\c^{1005}=z\end{matrix}\right.\) \(\Rightarrow x^2+y^2+z^2=xz+xz+yz\)
\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2xz+2yz\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-z=0\\y-z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z\)
\(\Rightarrow a^{1005}=b^{1005}=c^{1005}\Rightarrow a=b=c\)
\(\Rightarrow M=0\)
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bạn cũng xem phim gia sư siêu quậy reborn à ?
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Tính giá trị biểu thức:
\(\left(a-b\right)^{200}+\left(b-c\right)^{111}+\left(c-a\right)^{330}\)
biết \(a^{2010}+b^{2010}+c^{2010}=a^{1005}b^{1005}+b^{1005}c^{1005}+c^{1005}a^{1005}\)
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 + 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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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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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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