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Ta có \(a=\frac{2015b}{2016};c=\frac{2017b}{2016}\)
\(B=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(\frac{2015b}{2016}-b\right)\left(b-\frac{2017b}{2016}\right)-\left(\frac{2017b}{2016}-\frac{2015b}{2016}\right)^2\)
\(=4\left(-\frac{1b}{2016}\right)\left(-\frac{1b}{2016}\right)-\left(\frac{2b}{2016}\right)^2\)
\(=2^2\left(\frac{1b}{2016}\right)^2-\left(\frac{2b}{2016}\right)^2=\left(\frac{2b}{2016}\right)^2-\left(\frac{2b}{2016}\right)^2=0\)
thay a=b=c=0 vào B ta được B=0 vậy ta sẽ chứng minh B=0
Đặt \(\frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}=k\)
suy ra
\(\hept{\begin{cases}a=2015k\\b=2016k\\c=2017k\end{cases}}\)
vậy
\(B=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(B=4\left(2015k-2016k\right)\left(2016k-2017k\right)-\left(2017k-2015k\right)^2\)
\(B=4\left(-k\right)\left(-k\right)-\left(2k\right)^2\)
\(B=4k^2-4k^2\)
\(B=0\)
2.Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow a+b+c-a-b+c=0\)
\(\Rightarrow2c=0\)
\(\Rightarrow c=0\)
Vậy c=0
BT5: Ta có: f(1)=1.a+b=1 =>a+b=1 (1)
f(2)=2a+b=4 (2)
Trừ (1) cho (2) ta có: 2a+b-a-b=4-1 => a=3
Với a=3 thay vào (1) ta có: 3+b=1 => b=-2
Vậy a=3, b=-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)
Ta có:\(\frac{3-x}{2021}+\frac{2020-x}{2019}+\frac{4033-x}{2017}+\frac{6042-x}{2015}=10\)
\(\Leftrightarrow\frac{3-x}{2021}-1+\frac{2020-x}{2019}-2+\frac{4033-x}{2017}-3+\frac{6042-x}{2015}-4=0\)
\(\Leftrightarrow\frac{3-x-2021}{2021}+\frac{2020-x-4038}{2019}+\frac{4033-x-6051}{2017}+\frac{6042-x-8060}{2015}=0\)
\(\Leftrightarrow\frac{-2018-x}{2021}+\frac{-2018-x}{2019}+\frac{-2018-x}{2017}+\frac{-2018-x}{2015}=0\)
\(\Leftrightarrow-\left(2018+x\right)\left(\frac{1}{2021}+\frac{1}{2019}+\frac{1}{2017}+\frac{1}{2015}\right)=0\)
\(\Leftrightarrow2018+x=0.Do\frac{1}{2021}+\frac{1}{2019}+\frac{1}{2017}+\frac{1}{2015}>0\)
\(\Leftrightarrow x=-2018\)
V...
\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)
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!
theo bài ra ta có
\(\frac{a^{2015}}{b^{2017}+c^{2019}}=\frac{b^{2017}}{a^{2015}+c^{2019}}=\frac{c^{2019}}{a^{2015}+b^{2017}}\)
=>\(\frac{a^{2015}}{b^{2017}+c^{2019}}+1=\frac{b^{2017}}{a^{2015}+c^{2019}}+1=\frac{c^{2019}}{a^{2015}+b^{2017}}+1\)
=> \(\frac{a^{2015}+b^{2017}+c^{2019}}{b^{2017}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+b^{2017}}\)
=> b2017+ c2019 = -(a2015) (1)
=> a2015+ c2019= -(b2017) (2)
=> a2015+ b2017= -(c2019) (3)
thay 1, 2, 3 vào S ta có:
S = \(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}\)
=> S =\(\frac{-\left(a^{2015}\right)}{a^{2015}}+\frac{-\left(b^{2017}\right)}{b^{2017}}+\frac{-\left(c^{2019}\right)}{c^{2019}}\)
S = -1 + -1 + -1
S = -3
vậy S ko phụ thuộc vào giá trị a,b,c
=> b2017+c2019 = a2015+c2019=a2015+b2017
=> b2017 = a2015 = c2019
=>S=\(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}=\frac{2a^{2015}}{a^{2015}}+\frac{2b^{2017}}{b^{2017}}+\frac{2c^{2019}}{c^{2019}}=2+2+2=6\)
VẬY S ko phụ thuộc vào các giá trị của a,b,c
từ 2 trường hợp trên => giá trị của biểu thức S ko phụ thuộc vào giá trị của a,b,c (đpcm)
thanks you :)