\(Cho:\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
\(and........a\ne b\ne c........a,b,c\ne0\)
Tính \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Những câu hỏi liên quan
1) Cho frac{a-left(c-bright)}{b-c}+frac{b-left(a-cright)}{c-a}+frac{c-left(b-aright)}{a-b}3CM frac{a}{left(b-cright)^2}+frac{b}{left(c-aright)^2}+frac{c}{left(a-bright)^2}02) Cho frac{1}{a}+frac{1}{c}frac{1}{b-c}-frac{1}{a-b}và acne0; ane b; bne cCM frac{a}{c}frac{a-c}{b-c}
Đọc tiếp
1) Cho \(\frac{a-\left(c-b\right)}{b-c}+\frac{b-\left(a-c\right)}{c-a}+\frac{c-\left(b-a\right)}{a-b}=3\)
CM \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
2) Cho \(\frac{1}{a}+\frac{1}{c}=\frac{1}{b-c}-\frac{1}{a-b}\)và \(ac\ne0\); \(a\ne b\); \(b\ne c\)
CM \(\frac{a}{c}=\frac{a-c}{b-c}\)
cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b,c\ne0;b\ne c\right)\)) chứng minh rằng : \(\frac{a}{b}=\frac{a-c}{c-b}\)
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b,c\ne0,b\ne c\right)\).Chứng minh rằng\(\frac{a}{b}=\frac{a-c}{c-b}\)
Cho: \(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b},\right)\left(a,b,c\ne0,b\ne c\right)\) Chứng minh rằng: \(\frac{a}{b}=\frac{a-b}{c-b}\)
cmr nếu\(a\left(z+y\right)=b\left(z+x\right)=c\left(x+y\right);a\ne b\ne c\ne0\Rightarrow\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
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đề đúng thì giải giùm ik bạn ơi
Cho \(a\ne b\ne c\ne0\)và\(a+b+c=0\)Tính:
\(A=\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right).\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)\)
Từ \(a+b+c=0\) bạn tự chứng minh \(a^3+b^3+c^3=3abc\)
Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}\frac{\left(a-b\right)\left(c-a-b\right)}{ab}\)
\(=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự, ta có: \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)
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\(\)Cho \(a\ne b\ne c\ne0\)và \(\frac{a+b}{c}=\frac{b+C}{a}=\frac{c=a}{b}\).Tính gá trị của bieur thức
M=\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
*Nếu a + b + c = 0
\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
Thay vào M đc
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)
\(=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}\)
\(=-1\)
*Nếu \(a+b+c\ne0\)
Áp dụng t.c của dãy tsbn
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)
\(\Rightarrow a=b=c\)
Thay vào M đc
\(M=\left(1+\frac{a}{a}\right)\left(1+\frac{b}{b}\right)\left(1+\frac{c}{c}\right)=2.2.2=8\)
Vậy ..............
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Xem thêm câu trả lời
\(\frac{a}{c}=\frac{a-b}{b-c}\left(a;c\ne0;a\ne b;b\ne c\right)\)
\(Cmr:\frac{1}{a}+\frac{1}{a-b}=\frac{1}{b-c}-\frac{1}{c}\)
Biết \(a\ne-b,b\ne-c,c\ne-a\). CMR:
\(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}+\frac{-a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}\)