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\(\frac{a}{2b}\)=\(\frac{b}{2c}\) =\(\frac{c}{2d}\) =\(\frac{d}{2a}\)=\(\frac{a+b+c+d}{2a+2b+2c+2d}\)=\(\frac{a+b+c+d}{2\left(a+b+c+d\right)}\)=\(\frac{1}{2}\)
quên rùi............................
đáp số =2
Áp dụng TC DTSBN ta có :
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{a}{2b}=\frac{1}{2}\Rightarrow a=\frac{1}{2}.2b\Rightarrow a=b\) (1)
\(\Rightarrow\frac{b}{2c}=\frac{1}{2}\Rightarrow b=\frac{1}{2}.2c\Rightarrow b=c\) (2)
\(\Rightarrow\frac{c}{2a}=\frac{1}{2}\Rightarrow c=\frac{1}{2}.2a\Rightarrow c=a\) (3)
\(\Rightarrow\frac{d}{2a}=\frac{1}{2}\Rightarrow d=\frac{1}{2}.2a\Rightarrow d=a\) (4)
Từ (1);(2);(3):(4) \(\Rightarrow a=b=c=d\) .Thay vào A ta được :
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a+2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(=\frac{a}{2a}+\frac{4021a}{2a}+\frac{a}{2a}+\frac{a}{2a}=\frac{a+4021a+a+a}{2a}=\frac{4024a}{2a}=\frac{4024}{2}=2012\)
Vậy \(A=2012\)
a)\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
\(=>\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)
\(=>\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)
\(=>\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=2\) hoặc \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=-2\)
Bộ thứ1 (x,y,z)=(6,8,10)
Bộ thứ 2 (x,y,z)=(-6;-8;-10)
b) Theo đề bài \(=>\frac{2b}{a}=\frac{2c}{b}=\frac{2d}{c}=\frac{2a}{d}=\frac{2.\left(a+b+c+d\right)}{a+b+c+d}=2\)
=>a=b=c=d
\(=>A=\frac{2011a-2010a}{2a}.4=\frac{a}{2a}.4=2\)( thay b,c,d=a, vì a=b=c=d)
Áp dụng dãy tỉ số bằng nhau ta có: \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{1}{2}\)
\(\Rightarrow a=b;b=c;c=d;d=a\) hay \(a=b=c=d\)
\(\Rightarrow A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d+\right)}=\frac{1}{2}\)=\(\frac{1}{2}\)
\(\Rightarrow2a=2b,2b=2c,2c=2d,2d=2a\)
\(\Leftrightarrow a=b=c=d\)
\(\Rightarrow A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}=\frac{2011d-2010a}{b+c}\)
\(\Leftrightarrow A=\frac{2011a-2010a}{a+a}+\frac{2011b-2010b}{b+b}+\frac{2011c-2010c}{c+c}+\frac{2011d-2010d}{d+d}\)
\(\Leftrightarrow A=\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}+\frac{d}{2d}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
b) Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)
\(\Rightarrow x=15k,y=20k,z=24k\)
Lại có: \(M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{2\left(15k\right)+3\left(20k\right)+4\left(24k\right)}{3\left(15k\right)+4\left(20k\right)+5\left(24K\right)}=\frac{30k+60k+96k}{45k+80k+120k}\)
\(=\frac{186k}{245k}=\frac{186}{245}\)
Vậy \(M=\frac{186}{245}\)
b) Vì \(\frac{x}{3}\) = \(\frac{y}{4}\) và \(\frac{y}{5}\) = \(\frac{z}{6}\)
nên \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\)
Đặt \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\) = k
=> x = 15k; y = 20k và z = 24k
Thay vào M ta được:
M = \(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)
= \(\frac{30k+60k+96k}{45k+50k+120k}\)
= \(\frac{k\left(30+60+96\right)}{k\left(45+50+120\right)}\)
= \(\frac{k.186}{k.215}\) = \(\frac{186}{215}\)
Vậy M = \(\frac{186}{215}\).