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Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b-2017c}{c}=\frac{b+c-2017a}{a}=\frac{c+a-2017b}{b}\)
\(=\frac{a+b-2017c+b+c-2017a+c+a-2017b}{a+b+c}=\frac{-2015\left(a+b+c\right)}{a+b+c}=-2015\)
Do đó :
\(\frac{a+b-2017c}{c}=-2015\)\(\Leftrightarrow\)\(a+b=2c\) \(\left(1\right)\)
\(\frac{b+c-2017a}{a}=-2015\)\(\Leftrightarrow\)\(b+c=2a\) \(\left(2\right)\)
\(\frac{c+a-2017b}{b}=-2015\)\(\Leftrightarrow\)\(c+a=2b\) \(\left(3\right)\)
Thay (1), (2) và (3) vào \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\) ta được :
\(B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{8abc}{abc}=8\)
Vậy \(B=8\)
Chúc bạn học tốt ~
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Ta có \(\frac{\left(a+2017b\right)^2}{(b+2017c)^2}=\frac{a^2+2017b^2}{ac+2017c^2}=\frac{a^2+2017ac}{ac+2017c^2}=\frac{a.\left(a+2017c\right)}{c.\left(a+2017c\right)}=\frac{a}{c}\)
=> ĐPCM
Học tốt
.............
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Từ \(b^2=ac\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)(1)
Ta có: \(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}=\left(\frac{a}{b}\right)^2\left(đpcm\right)\)
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Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{2017c-a-b}{c}=\frac{2017b-a-c}{b}=\frac{2017a-b-c}{a}=\frac{\left(2017c-a-b\right)+\left(2017b-a-c\right)+\left(2017a-b-c\right)}{a+b+c}=\frac{2015.\left(a+b+c\right)}{a+b+c}=2015\)
\(\frac{2017c-a-b}{c}=2015\)\(\Rightarrow2017c-a-b=2015c\)\(\Rightarrow2c=a+b\)( 1 )
\(\frac{2017b-a-c}{b}=2015\)\(\Rightarrow2017b-a-c=2015b\)\(\Rightarrow2b=a+c\)( 2 )
\(\frac{2017a-b-c}{a}=2015\)\(\Rightarrow2017a-b-c=2015a\)\(\Rightarrow2a=b+c\)( 3 )
Từ ( 1 ), ( 2 ) và ( 3 ) \(\Rightarrow a=b=c\)
Vậy A = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right).\left(1+1\right).\left(1+1\right)=2^3=8\)
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b) Ta có: [tex]\frac{a^{2} + c^{2}}{b^{2} + a^{2}}[/tex]= [tex]\frac{bc + c^{2}}{b^{2} + bc}= \frac{c(b +c)}{b(b + c)}= \frac{c}{b}[/tex] (đpcm)
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Chứng minh (2016a-2017b)/(2017c+2018d)=(2016c-2017d)/(2017a+2018b) - Nguyễn Minh Hải
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b^2=ac
b^2+2017bc=ac+2017bc
b(b+2017c)=c(a+2017b)
b/c=(a+2017b)/(b+2017c)
(b/c)^2=((a+2017b)/(b+2017c))^2
b^2/c^2=(a+2017b)^2/(b+2017c)^2
thế b^2=ac ta có
ac/c^2=(a+2017b)^2/(b+2017c)^2
a/c=(a+2017b)^2/(b+2017c)^2
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Bài 1:
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^2+2017b^2}{c^2+2017d^2}=\dfrac{b^2k^2+2017b^2}{d^2k^2+2017d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2+2017b^2}{c^2+2017d^2}=\dfrac{ab}{cd}\)
Ta có :
\(b^2=ac\)\(\Rightarrow\)\(\frac{a}{b}=\frac{b}{c}\)\(\Rightarrow\)\(\frac{a}{b}=\frac{2017b}{2017c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\)\(\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)\(\left(1\right)\)
Lại có :
\(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{ab}{bc}=\frac{a}{c}\)\(\left(2\right)\)
Từ (1) và (2) suy ra :
\(\frac{a}{c}=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\)
Vậy ...
Chúc bạn học tốt ~
Ta có: \(b^2=ac\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a}{b}=\frac{2017b}{2017c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{2017b}{2017c}=\frac{a+2017b}{b+2017c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a+2017b}{b+2017c}\right)^2=\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}\left(1\right)\)
Ta lại có:
\(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{ab}{bc}=\frac{a}{c}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{\left(a+2017b\right)^2}{\left(b+2017c\right)^2}=\frac{a}{c}\)