chung minh a/b=c/d thi [a-b/c-d]^2017=a^2017+b^2017/c^2017+d^2017
Cho b2 = a*c, c2 = b*d (với b, c, d khác 0), (b+c khác 0), (b2017 + c2017 khác d2017). Chứng minh rằng a2017 + b2017 - c2017 / b2017 + c2017 - d2017 = (a + b- c)2017 / (b + c -d)2017.
Cho b2 = a*c, c2 = b*d (với b, c, d khác 0), (b+c khác 0), (b2017 + c2017 khác d2017). Chứng minh rằng a2017 + b2017 - c2017 / b2017 + c2017 - d2017 = (a + b- c)2017 / (b + c -d)2017.
Cho a,b,c,d là 4 số khác 0; biết \(\dfrac{a}{b}=\dfrac{c}{d}\).Chứng minh rằng \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)
Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=@\Leftrightarrow\left\{{}\begin{matrix}a=b@\\c=d@\end{matrix}\right.\)
khi đó: \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{b^{2017}@^{2017}+b^{2017}}{d^{2017}@^{2017}+d^{2017}}=\dfrac{b^{2017}\left(@^{2017}+1\right)}{d^{2017}\left(@^{2017}+1\right)}=\dfrac{b^{2017}}{d^{2017}}\)
\(\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\dfrac{\left(b@-b\right)^{2017}}{\left(d@-d\right)^{2017}}=\dfrac{\left[b\left(@-1\right)\right]^{2017}}{\left[d\left(@-1\right)\right]^{2017}}=\dfrac{b^{2017}}{d^{2017}}\)
Ta có điều phải chứng minh
Cho b2=a.c và c2=b.d(với b;c;d khác 0;b+c không bằng d;b2017+c2017ko bằng d2017(ko bằng có nghĩa là lớn hơn hoặc nhỏ hơn một sô)). Chứng minh rằng \(\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}-d^{2017}}\)=\(\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)
Ta có:
b2=a.c c2=b.d
\(\Rightarrow\frac{b}{c}=\frac{a}{b}\) \(\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) (1)
\(\Rightarrow\hept{\begin{cases}\left(1\right)=\frac{a^{2017}}{b^{2017}}=\frac{b^{2017}}{c^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}d^{2017}}\\\left(1\right)=\frac{a+b-c}{b+c-d}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\end{cases}}\)
\(\Rightarrow\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)
Vậy \(\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)
Ta có: \(b^2=a\cdot c\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)
\(c^2=b\cdot d\Rightarrow\frac{b}{c}=\frac{c}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{b^{2017}}{c^{2017}}=\frac{c^{2017}}{d^{2017}}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a^{2017}}{b^{2017}}=\frac{b^{2017}}{c^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}-d^{2017}}\)(3)
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b-c}{b+c-d}\)
\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)(4)
Từ (3) và (4) \(\Rightarrow\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}-d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)(đpcm)
Bài 1: Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{b^{2017}\cdot k^{2017}+d^{2017}\cdot k^{2017}}{b^{2017}+d^{2017}}=k^{2017}\)
\(\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}=\dfrac{\left(bk+dk\right)^{2017}}{\left(b+d\right)^{2017}}=k^{2017}\)
Do đó: \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
Cho a,b,c,d là 4 số khác 0; biết \(\frac{a}{b}=\frac{c}{d}\).Chứng minh rằng \(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)
vì \(\frac{a}{b}\)=\(\frac{c}{d}\)=>\(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\)
áp dụng tính chất dãy tỉ số bằng nhau
=> \(\frac{a^{2017}}{b^{2017}}\) =\(\frac{c^{2017}}{d^{2017}}\)= \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\)=\(\frac{a^{2017}-c^{2017}}{b^{2017}-d^{2017}}\)=\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)(diều phải chứng minh
Từ \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra a=bk
c=dk
Ta có
\(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(bk\right)^{2017}+b^{2017}}{\left(dk\right)^{2017}+d^{2017}}=\frac{b^{2017}.k^{2017}+b^{2017}}{d^{2017}.k^{2017}+d^{2017}}=\frac{b^{^{2017}}\left(k^{2017}+\right)}{d^{2017}\left(k^{2017}+1\right)}=\frac{b^{2017}}{d^{2017}}\)(1)
Ta có
\(\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\frac{\left(bk-b\right)^{2017}}{\left(dk-d\right)^{2017}}=\frac{\left(b\left(k-1\right)\right)^{2017}}{\left(d\left(k-1\right)\right)^{2017}}=^{\frac{b^{2017}}{d^{2017}}}\)(2)
Từ (1) và (2)
Ta suy ra
\(\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}\)
từ gt: \(\frac{a}{b}\)=\(\frac{c}{d}\)suy ra ad=bc
\(\frac{a^{2017}+b^{2017}=\left(a-b\right)^{2017}}{^{c^{2017}}+d^{2017}=\left(c-d\right)^{2017}}\)
suy ra \(a^{2017}+b^{2017}.\left(c-d\right)^{2017}=c^{2017}+d^{2017}.\left(a-b\right)^{2017}\)
\(a^{2017}+b^{2017}.c^{2017}-b^{2017}.d^{2017}=c^{2017}+d^{2017}.a^{2017}-d^{2017}.b^{2017}\)
theo mình nghĩ là\(b^{2017}.c^{2017}=d^{2017}.a^{2017}\)
bc=da
Cho a,b,c,d thuộc Z.Thỏa mãn a+b=c+d.Mà a2+b2=c2+d2.Chứng minh a^2017+b^2017=c^2017+d^2017
a^2017+b^2017/c^2017+c^2017=a^2017-b^2017/c^2017-d^2017
CM: a/b = c/d biết :
a mũ 2017 + b mũ 2017 a mũ 2017 - b mũ 2017
------------------------------- = ------------------------------
c mũ 2017 + d mũ 2017 c mũ 2017 - d mũ 2017
Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^{2017}=\left(\frac{b}{d}\right)^{2017}=\frac{a^{2017}}{c^{2017}}=\frac{b^{2017}}{d^{2017}}=\frac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\frac{a^{2017}-b^{2017}}{c^{2017}-d^{2017}}\left(đpcm\right)\)