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\(\frac{a-b}{1+ab}+\frac{b-c}{1+bc}+\frac{c-a}{1+ac}\)
\(=\frac{a-b}{1+ab}+\frac{b-a+a-c}{1+bc}+\frac{c-a}{1+ac}\)
\(=\frac{a-b}{1+ab}+\frac{b-a}{1+bc}+\frac{a-c}{1+bc}+\frac{c-a}{1+ac}\)
\(=\frac{b-a}{1+bc}-\frac{b-a}{1+ab}-\frac{c-a}{1+bc}+\frac{c-a}{1+ac}\)
\(=\left(b-a\right)\left(\frac{1}{1+bc}-\frac{1}{1+ab}\right)-\left(c-a\right)\left(\frac{1}{1+bc}-\frac{1}{1+ac}\right)\)
\(=\left(b-a\right)\left(\frac{1+ab-1-bc}{\left(1+ab\right)\left(1+bc\right)}\right)-\left(c-a\right)\left(\frac{1+ac-1-bc}{\left(1+bc\right)\left(1+ac\right)}\right)\)
\(=\left(b-a\right)\frac{b\left(a-c\right)}{\left(1+ab\right)\left(1+bc\right)}-\left(c-a\right)\frac{c\left(a-b\right)}{\left(1+bc\right)\left(1+ac\right)}\)
Quy đồng:
\(=\frac{\left(b-a\right)b\left(a-c\right)\left(1+ac\right)-\left(c-a\right)c\left(a-b\right)\left(1+ab\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)
\(=\frac{\left(b-a\right)b\left(a-c\right)\left(1+ac\right)-\left(a-c\right)c\left(b-a\right)\left(1+ab\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)
\(=\frac{\left(b-a\right)\left(a-c\right)\left(b\left(1+ac\right)-c\left(1+ab\right)\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)
\(=\frac{\left(b-a\right)\left(a-c\right)\left(b+abc-c-abc\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)là tích của chúng.
Ta có
\(\frac{a-b}{1+ab}=\frac{b-c}{1+bc}=\frac{a-c}{1+ac}\) nên
\(\frac{a-b}{1+ab}+\frac{b-c}{1+bc}+\frac{c-a}{1+ca}=\frac{a-b}{1+ab}+\frac{b-a}{1+bc}+\frac{a-c}{1+bc}+\frac{c-a}{1+ca}\)
\(=\left(a-b\right)\left[\frac{1}{1+ab}-\frac{1}{1+bc}\right]+\left(c-a\right)\left[\frac{1}{1+ac}-\frac{1}{1+bc}\right]\)
\(=\frac{\left(a-b\right)\left(1+bc-1+ab\right)}{\left(1+ab\right)\left(1+bc\right)}+\frac{\left(c-a\right)\left(1+bc-1-ac\right)}{\left(1+ac\right)\left(1+bc\right)}\)
\(=\frac{b\left(c-a\right)\left(a-b\right)}{\left(1+ab\right)\left(1+bc\right)}+\frac{c\left(c-a\right)\left(b-a\right)}{\left(1+ac\right)\left(1+bc\right)}\)
\(=\frac{\left(a-b\right)\left(c-a\right)}{\left(1+bc\right)}\left[\frac{b}{1+ab}-\frac{c}{1+ac}\right]\)
\(=\frac{\left(a-b\right)\left(c-a\right)\left(b-c\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\left(đpcm\right)\)
Bài 1:
Từ \(a+b+c=0\) ta có:
\(B=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-b^2-a^2}\)
\(=\frac{a^2}{(-b-c)^2-b^2-c^2}+\frac{b^2}{(-c-a)^2-c^2-a^2}+\frac{c^2}{(-b-a)^2-b^2-a^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
Lại có:
\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3\)
\(=-c^3+3abc+c^3=3abc\)
Do đó \(B=\frac{3abc}{2abc}=\frac{3}{2}\)
Bài 2:
Lấy P-Q ta có:
\(P-Q=\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)\)
\(P-Q=\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}\)
\(P-Q=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}+\frac{(b-c)(b^2+bc+c^2)}{b^2+bc+c^2}+\frac{(c-a)(c^2+ac+a^2)}{c^2+ac+a^2}\)
\(P-Q=(a-b)+(b-c)+(c-a)=0\Rightarrow P=Q\)
Ta có đpcm.
Bài 2:
Bài 1:
\(a^2+b^2+c^2=14\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=14\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=14\Rightarrow ab+bc+ac=-7\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có:
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=14^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=196-2.49=98\)
a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
Lời giải:
Có: \(\frac{a+b-c}{ab}-\frac{b+c-a}{bc}-\frac{c+a-b}{ca}=0\)
\(\Leftrightarrow \frac{c(a+b-c)}{abc}-\frac{a(b+c-a)}{abc}-\frac{b(c+a-b)}{abc}=0\)
\(\Leftrightarrow c(a+b-c)-a(b+c-a)-b(c+a-b)=0\)
Thực hiện khai triển và rút gọn:
\(\Leftrightarrow a^2+b^2-c^2-2ab=0\)
\(\Leftrightarrow (a-b)^2-c^2=0\)
\(\Leftrightarrow (a-b-c)(a-b+c)=0\) \(\Rightarrow \left[\begin{matrix} a=b+c\\ b=a+c\end{matrix}\right.\)
Tức là trong ba số $a,b,c$ có một số bằng tổng của hai số kia.
Ta có đpcm.
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
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
\(\left(a+b+c\right)=\dfrac{1}{2}\Leftrightarrow\left(a+b+c\right)^2=\dfrac{1}{4}\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=\dfrac{1}{4}\)
Ta có: \(ab+bc+ac=\left(a^2+b^2+c^2+2ab+2bc+2ac\right)-\left(a^2+b^2+c^2+ab+bc+ac\right)=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1}{12}\)
\(a^2+b^2+c^2=\dfrac{1}{6}-\left(ab+bc+ac\right)=\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
Suy ra: \(a^2+b^2+c^2=ab+bc+ac\Leftrightarrow a=b=c\)
\(P=\dfrac{3}{2}\)
p/s làm lih tih k chắc đâu:v
Ta có \(\dfrac{a-b}{ab+1}+\dfrac{b-c}{bc+1}+\dfrac{c-a}{ca+1}=\dfrac{\left(a-b\right)\left(bc+1\right)\left(ca+1\right)+\left(b-c\right)\left(ca+1\right)\left(ab+1\right)+\left(a-b\right)\left(bc+1\right)\left(ca+1\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\).