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Ta có: \(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{2017}=\frac{b}{2018}=\frac{c}{2019}=\frac{a-b}{2017-2018}=\frac{b-c}{2018-2019}=\frac{a-c}{2017-2019}.\)
\(\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{a-c}{-2}\)
\(\Rightarrow\frac{a-b}{-1}.\frac{b-c}{-1}=\left(\frac{a-c}{-2}\right)^2\)
\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(a-c\right)^2}{\left(-2\right)^2}\)
\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(a-c\right)^2}{4}.\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2.1\)
\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2\left(đpcm\right).\)
Chúc bạn học tốt!
A=a/2018-c +b/2018-a +c/2018-b
A= a/a+b + b/b+c + c/c+a
Nhận thấy: a/a+b< a/a+b+c; b/b+c<b/a+b+c; c/c+a<c/a+b+c
Do đó A= a/a+b + b/b+c + c/c+a < a/a+b+c + b/a+b+c + c/a+b+c = a+b+c/a+b+c=1
=>A>1(1)
áp dụng t/c:a/b<1=>a/b<a+n/b+n(a,b,n khác 0), ta có:
a/a+b < a+c/a+b+c ; b/b+c < b+a/b+c+a ; c/c+a < c+b/c+a+b
Do đó :A= a/a+b + b/b+c + c/c+a < a+c/a+b+c + b+a/a+b+c + c+b/a+b+c= 2(a+b+c)/a+b+c=2
=>A<2(2)
từ (1);(2)=>1<A<2=> A không thuộc Z=>ĐPCM. chúc bạn học tốt
Câu 1:
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}\)
b,Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{b}{d}\cdot\frac{a}{c}\Rightarrow\frac{a^2}{b^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ac}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Ta lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a+b}{c+d}\cdot\frac{a+b}{c+d}\Rightarrow\frac{ab}{cd}=\left(\frac{a+b}{c+d}\right)^2\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Câu 2:
\(\frac{a1}{a2}=\frac{a2}{a3}=....=\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+....+a2018}\)
\(\Rightarrow\frac{a1}{a2}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(1\right)\)
\(\frac{a2}{a3}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2\right)\)
..............
\(\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2017\right)\)
Nhân các vế (1),(2)....(2017) ta được:
\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\cdot\cdot\cdot\cdot\frac{a2017}{a2018}=\frac{a1}{a2018}=\left(\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\right)^{2017}\)
Vậy...
Câu 3:
\(x_2^2=x_1x_3\Rightarrow\frac{x1}{x2}=\frac{x2}{x3}\)
\(x_3^2=x_2x_4\Rightarrow\frac{x2}{x3}=\frac{x3}{x4}\)
\(x_4^2=x_3x_5\Rightarrow\frac{x3}{x4}=\frac{x4}{x5}\)
\(x_5^2=x_4x_6\Rightarrow\frac{x4}{x5}=\frac{x5}{x6}\)
Đến đây thfi làm giống câu 2
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)
Đặt \(\dfrac{a}{2017}=\dfrac{b}{2018}=\dfrac{c}{2019}=k\Rightarrow a=2017k;b=2018k;c=2019k\)
M = 4(2017k - 2018k)(2018k - 2019k) - (2019k - 2017k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2
= 0