Cho a,b,c là 3 số thực khác 0 sao cho:\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
Tính : B=\(\left(1+\dfrac{b}{a}\right).\left(1+\dfrac{a}{c}\right).\left(1+\dfrac{c}{b}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 11 2021
1.
Đặt \(\left(x;y;z\right)=\left(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}\right)\Rightarrow\left\{{}\begin{matrix}1-x=\dfrac{b}{b+a}\\1-y=\dfrac{c}{b+c}\\1-z=\dfrac{a}{a+c}\end{matrix}\right.\)
\(\Rightarrow xyz=\dfrac{1}{8}\\ xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\\ \Rightarrow xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)-xyz\\ \Rightarrow2xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)=\dfrac{1}{4}\\ \Rightarrow x+y+z=\dfrac{3}{4}+xy+yz+zx\)
\(\RightarrowĐpcm\)
4 tháng 12 2021
Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
16 tháng 11 2021
Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\)
\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{-abc}{abc}=-1\)
Với \(a+b+c\ne0\)
\(\dfrac{a+b-2021c}{c}=\dfrac{b+c-2021a}{a}=\dfrac{c+a-2021b}{b}=\dfrac{-2019\left(a+b+c\right)}{a+b+c}=-2019\\ \Leftrightarrow\left\{{}\begin{matrix}a+b-2021c=-2019c\\b+c-2021a=-2019a\\c+a-2021b=-2019b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)
16 tháng 11 2021
Với a+b+c=0⇔⎧⎪⎨⎪⎩b+c=−ac+a=−ba+b=−ca+b+c=0⇔{b+c=−ac+a=−ba+b=−c
B=a+ba⋅a+cc⋅b+cb=−abcabc=−1B=a+ba⋅a+cc⋅b+cb=−abcabc=−1
Với a+b+c≠0a+b+c≠0
a+b−2021cc=b+c−2021aa=c+a−2021bb=−2019(a+b+c)a+b+c=−2019⇔⎧⎪⎨⎪⎩a+b−2021c=−2019cb+c−2021a=−2019ac+a−2021b=−2019b⇔⎧⎪⎨⎪⎩a+b=2cb+c=2ac+a=2b
20 tháng 12 2020
Xét 2 TH sau:
TH1: a+b+c=0
Khi đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)
TH2: a+b+c khác 0
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Suy ra: a+b=2c; b+c=2a; c+a=2b
Do đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
20 tháng 12 2020
Xét 2 TH sau:
TH1: a+b+c=0
Khi đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)
TH2: a+b+c khác 0
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Suy ra: a+b=2c; b+c=2a; c+a=2b
Do đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
C
16 tháng 1 2023
`VT = (b-c)/((a-b)(a-c)) + (c-a)/((b-c)(b-a)) +(a-b)/((c-a)(c-b)) = 2/(a-b) + 2/(b-c) + 2/(c-a)`
`=-((a-b-a+c)/((a-b)(a-c))+(b-c-b+a)/((b-c)(b-a))+(c-a-c+b)/((c-a)(c-b)))`
`=-((a-b)/((a-b)(a-c))-(a-c)/((a-b)(a-c))+(b-c)/((b-c)(b-a))-(b-a)/((b-c)(b-a))+(c-a)/((c-a)(c-b))-(c-b)/((c-a)(c-b)))`
`= 1/(c-a)+1/(a-b)+1/(a-b)+1/(b-c)+1/(b-c)+1/(c-a)`
`=2/(a-b)+2/(b-c)+2/(c-a)=VP(đpcm)`
28 tháng 2 2021
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Ta có: \(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}=-1\)
Trường hợp 2: a=b=c
\(\Leftrightarrow N=\dfrac{b+b}{b}\cdot\dfrac{a+a}{a}\cdot\dfrac{c+c}{c}=2\cdot2\cdot2=8\)
OF
28 tháng 2 2021
1, Ta có a^3+b^3+c^3=3abc
-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2
-> (a+b)3 + c^3 - 3ab(a+b+c)=0
-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0
-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0
Th1: a+b+c=0
->P= a+b/2 . b+c/2 . c+a/2
= (-c)(-a)(-b)/2=-1
TH2 a^2+b^2+c^2-ab-bc-ca=0
->2a^2+2b^2+2c^2-2ab-abc-2ac=0
->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
-> (a-b)^2+(a-c)^2+(b-c)^2=0
Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0
Dấu = xảy ra (=)a-b=0
b-c=0
a-c=0
-> a=b=c
->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8
B= (1+a/b)(1+a/c)(1+c/b)
= (a/a+b/a)(c/c+a/c)(b/b+c/b)
= (a+b/a)(c+a/c)(b+c/b)
= a+b.c+a.b+c/acb
= a(b+c).(b+c)/acb
= (b+c)(b+c)/cb
= b(2c)/cb
= 2c/c
= 2