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Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
(a,b,c khác 0 nữa)
\(\dfrac{ab+1}{b}=\dfrac{bc+1}{c}=\dfrac{ca+1}{a}\)
\(\Leftrightarrow a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{c-b}{bc}\\b-c=\dfrac{a-c}{ca}\\c-a=\dfrac{b-a}{ab}\end{matrix}\right.\)(1)
Xét a=b hoặc b=c hoặc c=a thì=>a=b=c
Xét \(a\ne b\ne c\)
\(\left(1\right)\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)=\dfrac{\left(c-b\right)\left(a-c\right)\left(b-a\right)}{a^2b^2c^2}\)
\(\Leftrightarrow-1=\dfrac{1}{a^2b^2c^2}\)(vô nghiệm)
Vậy ...
Chắc bạn viết nhầm đề, cho \(a=b=c=1\) đâu có đúng
Sửa lại đề: cho \(abc=1\) chứng minh \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)
Ta có
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{ac+c+abc}\)
\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)
\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{ab+a+1}+\dfrac{1}{ab+a+1}\)
\(=\dfrac{a+ab+1}{ab+a+1}=1\) (đpcm)
mình nghĩ đề thế này, do bạn ko viết a+1,b+1,c+1 dưới mẫu
Cho abc = 1 . CMR : \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\)
GIẢI
Ta có : \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\)
\(=\frac{ab+a+1}{ab+a+1}=1\)
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{1}{b+1+bc}+\dfrac{1}{c+1+ac}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{abc+ac+abc.c}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{1+ac+c}+\dfrac{1}{ac+c+c}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac+1+c}{ac+c+1}=1\) (đpcm)
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{abc+ac+1}+\dfrac{ab}{abc+ab+1}+\dfrac{bc}{abc+bc+1}\)
\(=\dfrac{ac}{ac+2}+\dfrac{ab}{ab+2}+\dfrac{bc}{bc+2}\)
\(=abc\left(\dfrac{b}{abc+2}+\dfrac{c}{abc+2}+\dfrac{a}{abc+2}\right)\)
\(=1.1=1\)(đpcm).
Vậy \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\).