Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh: \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)(b,d ≠ 0; b≠ d). Chứng minh rằng : \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có
\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)
\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)
\(\Rightarrow VT=VP\)
Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)
cho \(\frac{a}{b}=\frac{c}{d}\left(b,d\ne0;b\ne d,-d\right)\)
Chứng minh \(\left(\frac{a-b}{c-d}\right)^{2018}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)
với c=0=>a=0 đẳng thức đúng
với c khác 0 ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{\left(a-b\right)^{2018}}{\left(c-d\right)^{2018}}=\frac{a^{2018}}{c^{2018}}=\frac{b^{2018}}{d^{2018}}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)
=>\(\frac{\left(a-b\right)^{2018}}{\left(c-d\right)^{2018}}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)
cho a,b,c là 3 cạnh tam giác
chứng minh
\(\frac{1}{\left(a+b-c\right)^{2018}}+\frac{1}{\left(a+c-b\right)^{2018}}+\frac{1}{\left(b+c-a\right)^{2018}}\ge\frac{1}{a^{2018}}+\frac{1}{b^{2018}}+\frac{1}{c^{2018}}\)
Trước tiên ta chứng minh bổ đề: Với x, y dương thì ta có:
\(\frac{1}{x^n}+\frac{1}{y^n}\ge\frac{2^{n+1}}{\left(x+y\right)^n}\)
Với n = 1 thì nó đúng.
Giả sử nó đúng đến \(n=k\)hay \(\frac{1}{x^k}+\frac{1}{y^k}\ge\frac{2^{k+1}}{\left(x+y\right)^k}\left(1\right)\)
Ta chứng minh nó đúng đến \(n=k+1\)hay \(\frac{1}{x^{k+1}}+\frac{1}{y^{k+1}}\ge\frac{2^{k+2}}{\left(x+y\right)^{k+1}}\left(2\right)\)
Từ (1) và (2) cái ta cần chứng minh trở thành:
\(\frac{1}{x^{k+1}}+\frac{1}{y^{k+1}}\ge\left(\frac{1}{x^k}+\frac{1}{y^k}\right)\frac{2}{\left(x+y\right)}\)
\(\Leftrightarrow\left(y-x\right)\left(y^{k+1}-x^{k+1}\right)\ge0\)(đúng)
Vậy ta có ĐPCM.
Áp dụng và bài toán ta được
\(2\left(\frac{1}{\left(a+b-c\right)^{2018}}+\frac{1}{\left(b+c-a\right)^{2018}}+\frac{1}{\left(c+a-b\right)^{2018}}\right)\ge\frac{2^{2019}}{2^{2018}.a^{2018}}+\frac{2^{2019}}{2^{2018}.b^{2018}}+\frac{2^{2019}}{2^{2018}.c^{2018}}\)
\(\Leftrightarrow\frac{1}{\left(a+b-c\right)^{2018}}+\frac{1}{\left(b+c-a\right)^{2018}}+\frac{1}{\left(c+a-b\right)^{2018}}\ge\frac{1}{a^{2018}}+\frac{1}{b^{2018}}+\frac{1}{c^{2018}}\)
cho a,b,c là 3 cạnh tam giác
chứng minh
\(\frac{1}{\left(a+b-c\right)^{2018}}+\frac{1}{\left(a+c-b\right)^{2018}}+\frac{1}{\left(b+c-a\right)^{2018}}\ge\frac{1}{a^{2018}}+\frac{1}{b^{2018}}+\frac{1}{c^{2018}}\)
a) Cho các số dương a,b,c,d; c khác d và \(\frac{a}{b}\)=\(\frac{c}{d}\). Chứng minh rằng : \(\frac{\left(a^{2018}+b^{2018}\right)^{2019}}{\left(c^{2018}+d^{2018}\right)^{2019}}\)=\(\frac{\left(a^{2019}-b^{2019}\right)^{2018}}{\left(c^{2019}-d^{2019}\right)^{2018}}\)
b) Cho biết |3x + 2y| + |5z - 7x| + \(\left(xy+yz+xz-500\right)^{2022}\)= 0 . Tính giá trị : \(A=\left(3x-y-z\right)^{2021}\)
Các bạn giải giúp mik nhé. Mik cần gấp lắm. Ai giải trc mik sẽ tick cho
các bạn tham khảo nhé
a, Cho \(a^{2018}+b^{2018}+c^{2018}=\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Tính \(P=\left(a-b\right)^{2018}+\left(b-c\right)^{2018}+\left(c-a\right)^{2018}\)
b, Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)và \(\frac{2}{ab}-\frac{1}{c^2}=9\)
Tính \(P=\left(a+2b+c\right)^{2018}\)
Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)
Mà đẳng thức trên xảy ra dấu =
\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)
Bài kia tí nghĩ nốt, khó v
Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)
Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)
Cho \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{b+d+a}=\frac{d}{a+b+c}\)
Tính \(A=\frac{a^{2018}}{b^{2018}}+\frac{b^{2018}}{c^{2018}}+\frac{c^{2018}}{d^{2018}}+\frac{d^{2018}}{a^{2018}}\)
Cho abc=1 và a,b,c đôi một khác nhau
Tính giá trị P=\(\frac{2018+2019a^3}{a\left(a-b\right)\left(a-c\right)}+\frac{2018+2019b^3}{b\left(b-a\right)\left(b-c\right)}+\frac{2018+2019c^3}{c\left(c-a\right)\left(c-b\right)}\)
Cho a, b, c thỏa mãn \(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}\). Chứng minh \(4\left(a-b\right)\left(c-d\right)=\left(c-a\right)^2\)
Đề có sai ko bạn sao lại c-d ?
Sửa đề : Cần chứng minh \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Đặt :\(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}=k\)
\(\Rightarrow\hept{\begin{cases}a=2017k\\b=2018k\\c=2019k\end{cases}}\)
Khi đó :
\(4\left(a-b\right)\left(b-c\right)=4\left(2017k-2018k\right)\left(208k-2019k\right)\)
\(=4\cdot\left(-k\right)\cdot\left(-k\right)=4k^2\)
\(\left(c-a\right)^2=\left(2019k-2017k\right)^2=\left(2k\right)^2=4k^2\)
Do đó : \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\) (đpcm)