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Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\Rightarrow a=2016k;b=2017k;c=2018k\)
\(\frac{a}{24}+\frac{b}{4}=\frac{c}{2018}\)
\(\Rightarrow\frac{2016k}{24}+\frac{2017k}{4}=\frac{2018k}{2018}\)
\(\Rightarrow84k+504,25k=k\)
\(\Rightarrow k=0\)
\(\Rightarrow a,b,c=0\)
1) ADTCDTSBN
có: \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-7}=\frac{x-y-z}{3-5+7}=\frac{20}{5}=4.\)
=> ...
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\)
\(\Rightarrow\dfrac{a}{b}=1\Rightarrow a=b\)
\(\dfrac{b}{c}=1\Rightarrow b=c\)
\(\dfrac{c}{d}=1\Rightarrow c=d\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow a^{20}.b^{17}.c^{2017}=d^{20}.d^{17}.d^{2017}=d^{2054}\)
đpcm
Tham khảo nhé~
Theo đề bài, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)
Áp dụng dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=1\Rightarrow a=b\\\frac{b}{c}=1\Rightarrow b=c\\\frac{c}{d}=1\Rightarrow c=d\end{cases}}\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow a^{20}.b^{17}.c^{2017}=d^{20}.d^{17}.d^{2017}=d^{2054}\)
\(\Rightarrowđpcm\)
1) x(x-2) + 3(x+5) + 4x -15 =0
=> x\(^2\) - 2x + 3x + 15 + 4x - 15 = 0
=> ( x\(^2\) -2x + 3x + 4x ) + 15 - 15 = 0
=> x \(^2\) -2x+3x+4x = 0
=> x(x-2+3+4)=0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2+3+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}}\)
2) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=2017\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=2017.2017\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=2017^2\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=2017^2\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{c}{a+b}\right)=2017^2\)
\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{c}{a+b}\right)=2017^2\)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2-3\)
xin lỗi mik xin đc sửa lại 3 dòng cuối vì mik ghi nhầm :
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{b}{a+c}\right)=2017^2\)
\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)=2017^2\)
\(\Rightarrow3+\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=2017^2\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2-3\)
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)