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Lớp 7 gì mà dễ ẹc :))
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Rightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a-5b=-3c\)
\(\Leftrightarrow a-4a=-3c\)
\(\Leftrightarrow-3a=-3c\)
\(\Rightarrow a=c\)
Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)
4. (3/4-81)(3^2/5-81)(3^3/6-81)....(3^6/9-81).....(3^2011/2014-81)
mà 3^6/9-81=0 => (3/4-81)(3^2/5-81)....(3^2011/2014-81)=0
\(\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Rightarrow\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{\left(2a-b\right)+\left(b-c+a\right)}{\left(a+b\right)+\left(2a-b\right)}=\frac{3a-c}{3a}=\frac{2}{3}\)
\(\Rightarrow2\times3a=3\times\left(3a-c\right)\)
\(\Rightarrow6a=9a-3c\)
\(\Rightarrow6a-9a=-3c\)
\(\Rightarrow-3a=-3c\)
\(\Rightarrow\frac{-3a}{-3}=\frac{-3c}{-3}\)
\(\Rightarrow a=c\)
\(\Rightarrow\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(5b+4a\right)^5}{\left(5b+4a\right)^2\left(a+3a\right)^3}=\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Rightarrow3\times\left(2a-b\right)=2\left(a+b\right)\)
\(\Rightarrow6a-3b=2a+2b\)
\(\Rightarrow6a-2a=3b+2b\)
\(\Rightarrow4a=5b\)
\(\Rightarrow b=\frac{4a}{5}\)
\(\Rightarrow\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}=\left(\frac{5\times\frac{4a}{5}+4a}{4a}\right)^3=\left(\frac{4a+4a}{4a}\right)^3\)
\(\Rightarrow\left(\frac{8a}{4a}\right)^3=2^3=8\)
Áp dụng tính chất hãy tỉ số bằng nhau ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow a+b=2c;b+c=2a;a+c=2b\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{b}{a}=\frac{a}{c}=\frac{c}{b}=1\)
\(\Rightarrow B=2.2.2=8\)
ta có: \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a-a+a+b+b-b-c+c+c}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)
nếu a+b+c =0
=> a =0-b-c => a = -(b+c)
b = 0-a-c => b = -(a+c)
c = 0-a-b => c = -(a+b)
thay vào \(B=\left(1+\frac{-\left(a+c\right)}{a}\right).\left(1+\frac{-\left(b+c\right)}{c}\right).\left(1+\frac{-\left(a+b\right)}{b}\right)\)
\(B=\left(\frac{a-\left(a+c\right)}{a}\right).\left(\frac{c-\left(b-c\right)}{c}\right).\left(\frac{b-\left(a+b\right)}{b}\right)\)
\(B=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}\)
\(B=-1\)
nếu a+b+c khác 0
mà \(\frac{a+b+c}{c+a+b}=\frac{a}{c}=\frac{b}{a}=\frac{c}{b}=1\Rightarrow a=b=c\)
=> \(B=\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right).\left(1+\frac{c}{b}\right)\)
\(B=\left(1+1\right).\left(1+1\right).\left(1+1\right)\)
\(B=2.2.2\)
\(B=8\)
KL: B= -1 hoặc B=8
Chúc bn học tốt !!!!
Gợi ý :
Bước 1 : Cộng 6 vào các hạng tử đã cho ở đề bài
Bước 2 : xét 2 TH :
TH1 : a + b + c = 0
TH2 : a + b + c khác 0
Chúc học tốt !!!!
