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Ta có
a + b + c = abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Ta có:a+b+c=abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(\frac{a}{c}=\frac{a-b}{b-c}\Rightarrow a\left(b-c\right)=c\left(a-b\right)\) (1)
\(\frac{1}{c}+\frac{1}{a-b}=\frac{a-b+c}{c\left(a-b\right)}\) (2)
\(\frac{1}{b-c}-\frac{1}{a}=\frac{a-b+c}{a\left(b-c\right)}\) (3)
\(Từ\left(1\right),\left(2\right),\left(3\right)\Rightarrow\)điều phải chứng minh
a) \(A=\frac{a^2}{cb}+\frac{b^2}{ca}+\frac{c^2}{ab}\)
\(A=\frac{a^2.a+b^2.b+c^2.c}{abc}\)
\(A=\frac{a^3+b^3+c^3}{abc}\left(1\right)\)
Ta lại có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(-c\right)\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(2\right)\)
Lấy (2) thay vào (1), ta được:
\(\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)
a) cho a+b+c=0a+b+c=0 và abc khác 0 Tính a2(a2−b2−c2)+b2(b2−c2−a2)+c2(c2−b2−a2)
b) B mình k biết
Có \(\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(a+c\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)Thay vào M đc
\(M=\frac{a^2}{2bc}+\frac{b^2}{2ca}+\frac{c^2}{2ab}\)\(\Leftrightarrow M=\frac{1}{2}\left(\frac{a^3+b^3+c^3}{abc}\right)\)
Tháy hơi sai đề rồi
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: \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Rightarrow ab+bc+ca=0\)
Chứng minh đẳng thức này mà áp dụng:
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Khi đó
\(M=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}\)
\(=\frac{\left(a^3b^3+b^3c^3+c^3a^3\right)}{abc}=\frac{3a^2b^2c^2}{abc}=3abc\) Do ab+bc+ca=0
Ta có : \(\frac{a}{b}=\frac{b}{c}\)
=> a.c = b2
Chứng minh tương tự : a.b = c2
b.c = a2
Thay vào biểu thức M, ta có:
M = \(\frac{a^4.b^{1996}.c^{18}}{\left(c^2\right)^{1009}}\)= \(\frac{a^4.b^{1996}.c^{18}}{\left(a.b\right)^{1009}}\)= \(\frac{a^4.b^{1996}.c^{18}}{a^{1009}.b^{1009}}\)= \(\frac{b^{987}.c^{18}}{a^{1005}}\)
= \(\frac{b^{987}.\left(c^2\right)^9}{a^{1005}}\)= \(\frac{b^{987}.\left(a.b\right)^9}{a^{1005}}\)= \(\frac{b^{996}}{a^{996}}\)= \(\frac{b^{996}}{\left(b.c\right)^{498}}\)= \(\frac{b^{498}}{c^{498}}\)
= \(\frac{b^{498}}{\left(a.b\right)^{498}}\)= \(\frac{b^{249}}{a^{249}}\)= \(\frac{b^{248}.b}{a^{248}.a}\)= \(\frac{b^{248}.b}{\left(b.c\right)^{124}.a}\)= \(\frac{b^{124}.b}{c^{124}.a}\)= \(\frac{b^{124}.b}{\left(a.b\right)^{62}.a}\)
= \(\frac{b^{62}.b}{a^{62}.a}\)= \(\frac{b^{62}.b}{\left(b.c\right)^{31}.a}\)= \(\frac{b^{31}.b}{c^{31}.a}\)= \(\frac{b^{30}.b^2}{c^{30}.a^2}\)= \(\frac{b^{30}.b^2}{\left(a.b\right)^{15}.a^2}\)= \(\frac{b^{15}.b^2}{a.^{15}a^2}\)
= \(\frac{b^{14}.b^3}{a.^{14}a^3}\)= \(\frac{b^{14}.b^3}{\left(b.c\right)^7.a^3}\)= \(\frac{b^7.b^3}{c^7.a^3}\)= \(\frac{b^6.b^4}{c^6.a^4}\)= \(\frac{b^6.b^4}{\left(b.a\right)^3.a^4}\)= \(\frac{b^3.b^4}{a^3.a^4}\)= \(\frac{b^2.b^5}{a^2.a^5}\)= \(\frac{b^2.b^5}{\left(b.c\right)^2.a^5}\)= \(\frac{b^5}{c^2.a^5}\)=
= \(\frac{b^4.b}{c^2.a^5}\)= \(\frac{b^4.b}{a.b.a^5}\)= \(\frac{b^4}{a^6}\)= \(\frac{b^4}{\left(a.b\right)^3}\)= \(\frac{b}{a^3}\)= \(\frac{b}{a^2.a}\)= \(\frac{b}{\left(b.c\right).a}\)= \(\frac{1}{a.c}\)
Vâng, xin chỉ coi đáp án và đừng làm theo phương pháp này, đi thi làm cách này chắc mất cả 60 phút đấy...