Cho a,b,c thỏa mãn: 2010a - b2 = 2010b - c2 = 2010c - a2 = 1. Tính \(B=\frac{2012a+b+c}{a}+\frac{2012b+c+a}{b}+\frac{2012c+a+b}{c}\)
cho a,b,c là các số dương thỏa mãn a+b+c=1006
chứng minh \(\sqrt{2012a+\frac{\left(b-c\right)^2}{2}}+\sqrt{2012b+\frac{\left(c-a\right)^2}{2}}+\sqrt{2012c+\frac{\left(a-b\right)^2}{2}}>2\)
Cho a/b=c/d.CM 2010a+2011b/2010c+2011d=2012a-2013b/2012c-2013d
Đề: \(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\)
TÍNH M= \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
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Từ \(\frac{2a+b+c+d}{a}=\frac{a-2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
=> \(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a+b+c+d=0 => a+b=-(c+d); b+c=-(a+d)
=> \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=-4\)
Nếu a+b+c+d \(\ne\)0 => a=c=b=d
=> \(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}=4\)
cho dãy tỉ số bằng nhau
\(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\)
tính \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
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Cho dãy tỉ số bằng nhau:
\(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\)
tính M =\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
cho dãy tỉ số bằng nhau
\(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\)
tính M=\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
Ta có: \(\frac{2012a+b+c+d}{a}-2011=\frac{a+2012b+c+d}{b}-2011=\frac{a+b+2012c+d}{c}-2011\)
\(=\frac{a+b+c+2012d}{d}-2011\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+) Xét \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(a+d\right);c+d=-\left(a+b\right);a+d=-\left(b+c\right)\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(=\frac{a+b}{-\left(a+b\right)}+\frac{b+c}{-\left(b+c\right)}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(d+a\right)}\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
+) Xét \(a+b+c+d\) khác 0 \(\Rightarrow a=b=c=d\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Vậy...
cho dãy tỉ số bằng nhau
\(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\)
Tính M= \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
Ta có:
\(\frac{2012a+b+c+d}{a}-2011=\frac{a+2012b+c+d}{b}-2011=\frac{a+b+2012c+d}{c}-2011\)\(=\frac{a+b+c+2012d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+) Xét \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(a+d\right);c+d=-\left(a+b\right);a+d=-\left(b+c\right)\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(=\frac{a+b}{-\left(a+b\right)}+\frac{b+c}{-\left(b+c\right)}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(d+a\right)}\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
+) Xét \(a+b+c+d\ne0\Rightarrow a=b=c=d\)
\(\Rightarrow a+b=c+d;b+c=a+d;c+d=a+b;a+d=b+c\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{d+a}\)
\(=1+1+1+1=4\)
Vậy ...
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Câu hỏi của Đào Ngọc Bảo Anh - Toán lớp 7 - Học toán với OnlineMath
Cho dãy tỉ số bằng nhau : \(\frac{2012a+b+c+d}{a}=\frac{a+2012b+c+d}{b}=\frac{a+b+2012c+d}{c}=\frac{a+b+c+2012d}{d}\).Tính giá trị biểu thức: M=\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
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họ bảo ko có đường dẫn
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