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vì a+b+c=0 => a+b= -c; b+c=-a; c+a=-b
(1+a/b)(1+b/c)(1+c/a)
=(a+b/b)(b+c/c)(a+c/a)
= (-c/b)(-a/c)(-b/a)
=-1
Thay a = -2 ; b = 1 ; c = 1 ( vì -2 + 1 + 1 = 0 )
Ta có : \(A=\left(1+\frac{-2}{1}\right)\left(1+\frac{1}{1}\right)\left(1+\frac{1}{-2}\right)\)
\(A=-1.2..\frac{1}{2}\)
\(A=-1\)
\(1\)
Ta có: a+b+c=0 => a+b=-c;b+c=-a;a+c=-b
=>\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)
Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath
=>\(\frac{a-b+c}{2b}+1=\frac{c-a+b}{2a}+1=\frac{a-c+b}{2c}+1\)
\(\Rightarrow\frac{a+b+c}{2b}=\frac{a+b+c}{2a}=\frac{a+b+c}{2c}\)
*TH1: nếu a+b+c=0 => a+b=-c; b+c=-a; c+a=-b
=>P=\(\left(\frac{b+c}{b}\right)\left(\frac{a+b}{a}\right)\left(\frac{c+a}{c}\right)\)
=\(\frac{-a}{b}.\frac{-c}{a}.\frac{-b}{c}=\frac{-\left(a.b.c\right)}{a.b.c}=-1\)
*TH2: Nếu a+b+c khác 0: thì a=b=c
Khi đó P=2.2.2=8
Vậy P= -1 hoặc 8
P/s: Bài toán này khá hay đó !!
Ta có : \(a\left(\frac{1}{b}+\frac{1}{c}\right)=b\left(\frac{1}{a}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{a^2c+a^2b}{abc}=\frac{b^2c+ab^2}{abc}=\frac{c^2b+c^2a}{abc}\)
Mà : \(a,b,c>0\)
\(\Rightarrow a^2c+a^2b=b^2c+ab^2=c^2b+c^2a\)
+) Xét : \(a^2c+a^2b=b^2c+ab^2\)
\(\Leftrightarrow ab\left(a-b\right)+c\left(a^2-b^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(ab+ca+cb\right)=0\)
\(\Leftrightarrow a-b=0\Leftrightarrow a=b\) (1)
( Do \(a,b,c>0\Rightarrow ab+ca+cb>0\) )
+) Xét \(b^2c+ab^2=c^2b+c^2a\)
\(\Leftrightarrow bc\left(b-c\right)+a\left(b^2-c^2\right)=0\)
\(\Leftrightarrow\left(b-c\right)\left(bc+ab+ac\right)=0\)
\(\Leftrightarrow b-c=0\Leftrightarrow b=c\)(2)
( Do \(a,b,c>0\Rightarrow ab+ca+cb>0\) )
Từ (1) và (2) \(\Rightarrow a=b=c\) (đpcm)
\(\frac{a-b+c}{2b}=\frac{c-a+b}{2a}=\frac{a-c+b}{2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
=> 2a-2b+2c=2b <=> a+c=2b. Chia cả 2 vế cho c ta được: \(1+\frac{a}{c}=\frac{2b}{c}\)
Tương tự: \(1+\frac{c}{b}=\frac{2a}{b}\) và \(1+\frac{b}{a}=\frac{2c}{a}\)
=> \(\left(1+\frac{c}{b}\right)\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)=\frac{2a}{b}.\frac{2c}{a}.\frac{2b}{c}=\frac{8.abc}{abc}=8\)
Đáp số: 8
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
<=> \(\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1\)
<=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
<=> a + b + c = 0 hoặc a = b = c.
Th1: a + b + c = 0
=> a + b = - c ; a + c = -b ; b + c = -a.
Thế vào P :
\(P=\left(1+\frac{a}{b}\right)\cdot\left(1+\frac{b}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)
\(=\left(\frac{a+b}{b}\right)\cdot\left(\frac{b+c}{c}\right)\cdot\left(\frac{c+a}{a}\right)\)
\(=-\frac{c}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)
TH2: a = b = c. THế vào P
\(P=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
Vậy: P = -1 nếu a + b + c = 0
hoặc P = 8 nếu a = b = c.
\(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)
Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)\(\Rightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1=\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow P=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{\left(-a\right).\left(-b\right).\left(-c\right)}{abc}=-1\)
TH2: Nếu \(a+b+c\ne0\)\(\Rightarrow a=b=c\)
\(\Rightarrow\hept{\begin{cases}a+b=2b\\b+c=2c\\c+a=2a\end{cases}}\)\(\Rightarrow P=\frac{2b}{b}.\frac{2c}{c}.\frac{2a}{a}=2.2.2=8\)
Vậy \(P=-1\)hoặc \(P=8\)
Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )
Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )
Khó quá do anh thien
\(A=3+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)