Cho a/b=c/d. Chứng minh a^2021-b^2021/a^2021+b^2021=c^2021-d^2021/c^2021+d^2021
Những câu hỏi liên quan
Cho a,b , c,d thỏa mãn:
a/b=c/d ( a khác + b , c khác + d )
Chứng minh
(a-b/c-d)^2021=a^2021+b^2021/c^2021+d^2021
Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
=> \(\left(\frac{a}{c}\right)^{2021}=\left(\frac{b}{d}\right)^{2021}=\left(\frac{a-b}{c-d}\right)^{2021}\)
=> \(\frac{a^{2021}}{c^{2021}}=\frac{b^{2021}}{d^{2021}}=\left(\frac{a-b}{c-d}\right)^{2021}=\frac{a^{2021}+b^{2021}}{c^{2021}+d^{2021}}\)
=>\(\left(\frac{a-b}{c-d}\right)^{2021}=\frac{a^{2021}+b^{2021}}{c^{2021}+d^{2021}}\)(đpcm)
Cho b^2=ac (b+c khác 0)
Chứng minh: $\frac{ (a+b)^{2021} }{ (b+c)^{2021} }$=$\frac{ a^{2021}+ b^{2021} }{b^{2021}+c^{2021}}$
Cho a,b,c,d là các số thực dương thỏa mãn a+b=c+d và a^2+b^2=c^2+d^2.Tính a^2021 + b^2021 = c^2021+d^2021
Cho \(b^2\)=ac
Chứng minh: \(\dfrac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}\) = \(\dfrac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}\)
Cho a, b, c ≠ 0 thoả mãn \(\left\{{}\begin{matrix}a+b+c=2021\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2021}\end{matrix}\right.\) . Chứng minh: \(\frac{1}{a^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{a^{2021}+b^{2021}+c^{2021}}\)
Cho (a + b + c)(ab +bc + ca) = abc. Chứng minh rằng
a2021 + b2021 + c2021 = (a + b + c)2021
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c\left(ab+bc+ca\right)-abc=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c\left(bc+ca\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
- Với \(a=-b\Rightarrow a^{2021}=-b^{2021}\Rightarrow\left\{{}\begin{matrix}a^{2021}+b^{2021}+c^{2021}=c^{2021}\\\left(a+b+c\right)^{2021}=c^{2021}\end{matrix}\right.\)
\(\Rightarrow a^{2021}+b^{2021}+c^{2021}=\left(a+b+c\right)^{2021}\)
Hai trường hợp sau hoàn toàn tương tự
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Cho b^2=a*c b+c khác 0(a+b)^2021/(b+c)^2021=a^2021+b^2021/b^2021+c^2021
Cho a,b,c khác 0 và 1/a+1/b+1/c=1/(a+b+c)
Tính A=(a^2021+b^2021+c^2021)(1/a^2021+1/b^2021+1/c^2021)
CMR: Nếu: \(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\) thì: \(\dfrac{x^{2021}+y^{2021}+z^{2021}}{a^{2021}+b^{2021}+c^{2021}}=\dfrac{x^{2021}}{a^{2021}}+\dfrac{y^{2021}}{b^{2021}}+\dfrac{z^{2021}}{c^{2021}}\)
Ta thấy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\ge\dfrac{x^2}{a^2+b^2+c^2}+\dfrac{y^2}{a^2+b^2+c^2}+\dfrac{z^2}{a^2+b^2+c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\).
Mà đẳng thức xảy ra nên ta phải có x = y = z = 0 (Do \(a^2,b^2,c^2>0\)).
Thay vào đẳng thức cần cm ta có đpcm.
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