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TH1: a+b+c khác 0(a,b,c khác 0)
ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(2+\frac{a+b-c}{c}=2+\frac{b+c-a}{a}=2+\frac{c+a-b}{b}\)
\(\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)
\(\Rightarrow a=b=c\)
thay a=b=c vào M ta có:
\(M=\left(1+\frac{a}{a}\right).\left(1+\frac{c}{c}\right).\left(1+\frac{b}{b}\right)\)
\(\Rightarrow M=\left(1+1\right).\left(1+1\right).\left(1+1\right)=2^3=8\)
TH2: a+b+c=0(a,b,c khác 0)
\(\Rightarrow a=-\left(b+c\right)\)
\(\Rightarrow b=-\left(c+a\right)\)
\(\Rightarrow c=-\left(b+a\right)\)
thay các giá trị của a,b,c vào bt M ta có:
\(M=\left(1+\frac{-c-a}{a}\right).\left(1+\frac{-b-a}{b}\right).\left(1+\frac{-c-b}{c}\right)\)
\(M=\frac{-c}{a}.\left(-\frac{a}{b}\right).\left(-\frac{b}{c}\right)\)
\(M=-1\)
vậy ...
p/s: bài này cô linh giải đúng rồi, nhưng ngắn quá => hơi khó hiểu
còn bn Darwin Watterson thiếu 1 trường hợp
\(^{\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=\frac{a+b+c}{a+b+c}=1}\)
Ta có: \(\frac{a+b-c}{c}=1\Rightarrow a+b-c=c\)
\(\Rightarrow a+b=2c\)
\(\frac{b+c-a}{a}=1\Rightarrow b+c-a=a\)
\(b+c=2a\)
\(\frac{c+a-b}{b}=1\Rightarrow c+a-b=b\)
\(\Rightarrow c+a=2b\)
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)( Quy đồng lên như bình thường )
Thay từ trên vào biểu thức, ta có: \(=\frac{2a.2b.2c}{abc}=\frac{8\left(abc\right)}{abc}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)
CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)
Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Thay vào biểu thwusc M ta được M=3abc (ĐPCM)
2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó
Nếu không thấy thì em có thể quy đồng lên mà rút gọn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(b-a\right)-\left(c-a\right)}{\left(b-a\right)\left(c-a\right)}+\frac{\left(c-b\right)-\left(a-b\right)}{\left(c-b\right)\left(a-b\right)}+\frac{\left(a-c\right)-\left(b-c\right)}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{1}{c-a}-\frac{1}{b-a}+\frac{1}{a-b}-\frac{1}{c-b}+\frac{1}{b-c}-\frac{1}{a-c}\)
\(=\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=VP\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
\(\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1\)
\(\Leftrightarrow\frac{a+b}{c}+\frac{c}{c}=\frac{b+c}{a}+\frac{a}{a}=\frac{a+c}{b}+\frac{b}{b}\)
\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)
\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 9x2 - 36
=(3x)2-62
=(3x-6)(3x+6)
=4(x-3)(x+3)
b) 2x3y-4x2y2+2xy3
=2xy(x2-2xy+y2)
=2xy(x-y)2
c) ab - b2-a+b
=ab-a-b2+b
=(ab-a)-(b2-b)
=a(b-1)-b(b-1)
=(b-1)(a-b)
P/s đùng để ý đến câu trả lời của mình
dễ!Ta có:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)
Chứng minh tương tự,Ta được:
\(\hept{\begin{cases}\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\\\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\end{cases}}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)\(\Rightarrow\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=\frac{2013}{2}\)
Xong!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
\(\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)
\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :\(\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)\)
\(=\frac{ab-1}{b}.\frac{bc-1}{c}.\frac{ac-1}{a}\)
Ta lại có : \(\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)\left(c-\frac{1}{c}\right)\)
\(=\frac{a^2-1}{a}.\frac{b^2-1}{b}.\frac{c^2-1}{c}\)