Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b)Ta có: \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow\frac{a^2}{ab}+\frac{b^2}{ab}\ge2\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge2\left(đpcm\right)\)
\(a^5-a=a\left(a^4-1\right)\)
\(=a\left(a^2+1\right)\left(a^2-1\right)\)
\(=a\left(a^2+1\right)\left(a-1\right)\left(a+1\right)\)
\(=a\left(a^2-4+5\right)\left(a-1\right)\left(a+1\right)\)
\(=a\left(a^2-4\right)\left(a-1\right)\left(a+1\right)+5a\left(a+1\right)\left(a-1\right)\)
\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a+1\right)\left(a-1\right)\)
Tích 5 số nguyên liên tiếp chia hết cho 5 nên \(a^5-a⋮5\)
Ta có:\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}.2=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{b}{a.b}+\frac{a}{a.b}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{a.b}\)
\(\Rightarrow2.a.b=c\left(a+b\right)\)
\(\Rightarrow a.b+a.b=ca+cb\)
\(\Rightarrow ab-cb=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
hok tốt!!
a)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow\left(1-\frac{b}{a}\right)=\left(1-\frac{d}{c}\right)\)
\(\Leftrightarrow\frac{a-b}{a}=\frac{c-d}{c}\Leftrightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
b)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được;
\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{c+d}\)
c)
\(\frac{b}{a}=\frac{d}{c}\Leftrightarrow3+\frac{b}{a}=3+\frac{d}{c}\Leftrightarrow\frac{3a+b}{a}=\frac{3c+d}{c}\Leftrightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}.2=\frac{a}{ab}+\frac{b}{ab}\)
\(\Rightarrow2c=\frac{a+b}{ab}\)
\(\Rightarrow2ab=\left(a+b\right)c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-bc\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
với a,b,c khác 0 và b khác c
đpcm.
a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)
\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)
A/d tc dãy tỉ số bằng nhau
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow a=b=c\)
\(\Rightarrow a=b=c=2014\)
\(2014-\frac{2}{19}.2014+\frac{5}{33}.2014=...\)