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a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
=>\(\frac{a-b+c}{2b}+1=\frac{c-a+b}{2a}+1=\frac{a-c+b}{2c}+1\)
\(\Rightarrow\frac{a+b+c}{2b}=\frac{a+b+c}{2a}=\frac{a+b+c}{2c}\)
*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
\(\frac{a-b+c}{2b}=\frac{c-a+b}{2a}=\frac{a-c+b}{2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
=> 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
TH1:
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)
Khi đó:\(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)
\(=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}\)
\(=-1\)
\(TH2:a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a}{b+2c}=\frac{b}{c+2a}=\frac{c}{a+2b}=\frac{a+b+c}{3\left(a+b+c\right)}=\frac{1}{3}\)
\(\Rightarrow\hept{\begin{cases}3a=b+2c\\3b=c+2a\\3c=a+2b\end{cases}}\Rightarrow3\left(a+b+c\right)=3\left(a+b+c\right)\Rightarrow a=b=c\)
Khi đó:\(P=\left(1+\frac{a}{a}\right)\left(1+\frac{a}{a}\right)\left(1+\frac{a}{a}\right)=8\)
Ta có: \(\frac{2a+b+c}{a}=\frac{a+2b+c}{b}=\frac{a+b+2c}{c}\)
\(\Rightarrow\frac{2a+b+c}{a}-1=\frac{a+2b+c}{b}-1=\frac{a+b+2c}{c}-1\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Mà \(a,b,c\ne0\)
=> a = b= c
\(A=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}\)
\(=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}\)
\(=2+2+2=6\)
Đặt ab = x, bc = y, ca = z (x, y, z ≠ 0 thỏa mãn x^3 + y^3 + z^3 = 3xyz)
⇔ (x+y)^3 − 3xy(x + y) + z^3 = 3xyz <=> (x+y)^3 − 3xy(x + y) + z^3 = 3xyz
⇔ (x + y)^3 + z^3 − 3xy(x + y+ z) = 0 ⇔ (x + y)^3 + z^3 − 3xy(x + y + z) = 0
⇔ (x + y + z)[(x + y)^2 − z (x + y) + z^2] − 3xy(x + y + z) = 0 ⇔ (x + y + z)[(x + y)^2 − z(x + y) + z2] − 3xy(x + y + z) = 0
⇔ (x + y + z)(x^2 + y^2 + z^2 − xy − yz − xz) = 0 ⇔ (x + y + z)(x^2 + y^2 + z^2 − xy − yz − xz) = 0
<=> x + y + z = 0 (1) và x^2 + y^2 + z^2 − xy − yz − xz = 0 (2)
Với (1): ⇔ ab + bc + ac = 0 ⇔ ab + bc + ac = 0
P = (1 + a/b)(1 + b/c)(1 + c/a) = (a + b)(b + c)(c + a)/abc=(ab + bc + ac)(a + b + c) − abc/abc = 0 − abc/abc = −1
Với (2) ⇔ (x − y)^2 + (y − z)^2 + (z − x)^2/2 = 0
⇔ (x − y)^2 + (y − z)^2 + (z − x)^2 = 0
Ta thấy (x − y)^2; (y − z)^2; (z − x)^2 ≥ 0 ∀x, y, z nên để tổng của chúng bằng 0 thì:
(x − y)^2 = (y − z)^2 = (z − x)^2 = 0 ⇒ x = y = z
⇔ ab = bc = ac ⇔ a=b=c (do a, b, c ≠ 0)
⇒ A = (1 + 1)(1 + 1)(1 + 1) = 8
Vậy...........
+) x + b + c ≠ 0
Ta có :
\(\frac{a-b+c}{2b}=\frac{c-a+b}{2a}=\frac{a-c+b}{2c}\)
\(\Rightarrow\frac{a-b+c}{2b}+1=\frac{c-a+b}{2a}+1=\frac{a-c+b}{2c}+1\)
\(\Rightarrow\frac{a+b+c}{2b}=\frac{a+b+c}{2a}=\frac{a+b+c}{2c}\)=> 2a = 2b = 2c ( do a + b + c ≠ 0 )
\(\Rightarrow a=b=c\Rightarrow P=\left(1+\frac{c}{c}\right).\left(1+\frac{b}{b}\right).\left(1+\frac{a}{a}\right)=2.2.2=8\)
+) a + b + c = 0
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a-b+c}{2b}=\frac{c-a+b}{2a}=\frac{a-c+b}{2c}=\frac{a-b+c+c-a+b+a-c+b}{2b+2a+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{0}{0}\left(\text{vô lý}\right)\)
Vậy P chỉ nhận 1 giá trị là P = 8