Cho:\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)(a,b,c,d>0)
Cho \(\frac{a+2c}{b+2d}=\frac{2a+c}{2b+d}\) .
CMR : \(\frac{a}{b}=\frac{a+c}{b+d};\frac{2a-c}{2b-d}=\frac{a-2c}{b-2d};\frac{a+2b}{a-b}=\frac{c+2d}{c-d}\)
Cho:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)
Tính: P\(\frac{2a-b}{2c-d}+\frac{2b-c}{2d-a}+\frac{2c-d}{2a-b}+\frac{2d-a}{2b-c}\)
Giúp với ai nhanh mình tick cho.
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=> a = b = c = d
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
Cho a,b,c,d>0
Cho \(A=\frac{2a+b+c}{a+b+c}+\frac{2b+c+d}{b+c+d}+\frac{2c+d+a}{c+d+a}+\frac{2d+a+b}{d+a+b}\)
Tìm [A]
Cho:\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)tinh:\frac{2011a-2010b}{c+d}=\frac{2011b-2010a}{c+d}=\frac{2011c+2011d}{a+b}=\frac{2011d-2010a}{c+dc=d}\)
cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\) trong đó a + b +c +d khác 0.tính giá trị biểu thức \(\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}\)
dễ mà bạn k cho mình thì mình giải cho k đi ko thì bye
cho a,b,c,d > 0. CMR \(\frac{a^4}{a^3+2b^3}+\frac{b^4}{b^3+2c^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\)
Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)\)
Tính \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}=\frac{2011d-2010a}{b+c}\)
\(\frac{a}{2b}\)=\(\frac{b}{2c}\) =\(\frac{c}{2d}\) =\(\frac{d}{2a}\)=\(\frac{a+b+c+d}{2a+2b+2c+2d}\)=\(\frac{a+b+c+d}{2\left(a+b+c+d\right)}\)=\(\frac{1}{2}\)
quên rùi............................
đáp số =2
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 \(A=\frac{2011a-2011b}{c+d}+\frac{2011b-2011c}{a+d}+\frac{2011c-2011d}{a+b}+\frac{2011d-2011a}{b+c}\)
Ta có: \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\) (ĐK: a,b,c,d > 0)
Theo đề bài, suy ra: \(\frac{2b}{a}=\frac{2c}{b}=\frac{2d}{c}=\frac{2a}{d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow\frac{2011a-2010a}{2a}.4=\frac{a}{2a}.4=2\) (Thay b, c ,d = a , Vì a = b =c =d)