cho tỉ lệ thức a/b=c/d.chứng minh rằng 2012a+2013b/2012a-2013b=2012c+2013d/2012c2013d
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Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng rằng \(\frac{2012a+2013b}{2012a-2013b}=\frac{2012c+2013d}{2012c-2013d}\)
Ta có :
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{2012a}{2012c}=\frac{2013b}{2013d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\frac{2012a}{2012c}=\frac{2013b}{2013d}=\frac{2012a+2013b}{2012c+2013d}=\frac{2012a-2013b}{2012c-2013d}\)
\(\Rightarrow\frac{2012a+2013b}{2012a-2013b}=\frac{2012c+2013d}{2012c-2013d}\)
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Cho a/b=c/d.CM 2010a+2011b/2010c+2011d=2012a-2013b/2012c-2013d
Cho\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\left(a,b,c,d>0\right)\)
Tính A=\(\dfrac{2013a-2012b}{c+d}+\dfrac{2013b-2012c}{a+d}+\dfrac{2013c-2012d}{a+b}+\dfrac{2013d-2012a}{b+c}\)
Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\) (a, b, c, d > 0). Tính:
A=\(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
cho\(\frac{a}{2b}\)=\(\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)(a, b, c, d > 0). Tính:
A=\(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
Vì a ; b ; c ; d > 0
=> a + b + c + d > 0
=> 2(a + b + c + d) > 0
=> 2a + 2b + 2c + 2d > 0
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
=> \(\frac{a}{2b}=\frac{1}{2}\Rightarrow2a=2b\Rightarrow a=b\)
Tương tự,ta được a = b = c = d
Khi đó A = \(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
= \(\frac{2013a-2012a}{2a}+\frac{2013b-2012b}{2b}+\frac{2013c-2012c}{2c}+\frac{2013d-2012d}{2d}\)(Vì a = b = c = d)
= \(\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}+\frac{d}{2d}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
\(a,b,c,d>0\text{ nên : }a+b+c+d>0\Rightarrow\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
do đó: a=b=c=d hay A=1/2+1/2+1/2+1/2=2
Tìm x biết
a)\(||3x-\frac{7}{3}|-2|=7\)
b) Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)(a, b, c, d > 0). Tính
A = \(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
Bài giải
a, \(\left| |3x-\frac{7}{3} | -2\right|=7\)
\(\Rightarrow\orbr{\begin{cases}|3x-\frac{7}{3}|-2=-7\\|3x-\frac{7}{3}|-2=7\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}|3x-\frac{7}{3}|=-5\text{ ( loại) }\\|3x-\frac{7}{3}|=9\end{cases}}\) \(\Rightarrow\text{ }\left|3x-\frac{7}{3}\right|=9\) \(\Rightarrow\orbr{\begin{cases}3x-\frac{7}{3}=-9\\3x-\frac{7}{3}=9\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}3x=\frac{-20}{3}\\3x=\frac{34}{3}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{20}{9}\\x=\frac{34}{9}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-\frac{20}{9}\text{ ; }\frac{34}{9}\right\}\)
CHO frac{a}{2b}frac{b}{2c}frac{c}{2d}frac{d}{2a} (a,b,c,d 0).TÍNH:Afrac{2013a-2012b}{c+d}+frac{2013b-2012c}{a+d}+frac{2013c-2012d}{a+b}+frac{2013d-2012a}{b+c}
Đọc tiếp
CHO \(\frac{a}{2b}\)=\(\frac{b}{2c}\)=\(\frac{c}{2d}\)=\(\frac{d}{2a}\) (a,b,c,d > 0).TÍNH:
A=\(\frac{2013a-2012b}{c+d}\)+\(\frac{2013b-2012c}{a+d}\)+\(\frac{2013c-2012d}{a+b}\)+\(\frac{2013d-2012a}{b+c}\)
Chứng minh 2013 < \(\dfrac{2013a}{a+b+c}+\dfrac{2013b}{b+c+d}+\dfrac{2013c}{c+d+a}+\dfrac{2013d}{d+a+b}< 4026\)
11 tháng 8 2023 lúc 15:40
Cho a,b,c là các số không âm thoả mãn a+b+c2006
Chứng minh rằng :
sqrt{2012a+dfrac{left(b-cright)^2}{2}}+sqrt{2012b+dfrac{left(c-aright)^2}{2}}+sqrt{2012c+dfrac{left(a-cright)^2}{2}}≤2012sqrt{2}
Đọc tiếp
Cho \(a,b,c\) là các số không âm thoả mãn \(a+b+c=2006\)
Chứng minh rằng :
\(\sqrt{2012a+\dfrac{\left(b-c\right)^2}{2}}\)\(+\)\(\sqrt{2012b+\dfrac{\left(c-a\right)^2}{2}}\)\(+\)\(\sqrt{2012c+\dfrac{\left(a-c\right)^2}{2}}\)≤\(2012\sqrt{2}\)