Cho \(abc\ne0\) và \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
\(Cho\)\(abc\ne0,và\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right).\)
Cho abc \(\ne0\) và \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính P = \(\left(1-\frac{b}{a}\right)\left(1+\frac{c}{d}\right)\left(1+\frac{a}{c}\right)\)
Cho \(abc\ne0\) và \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{a+c-b}{b}+2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Nếu a+b+c=0
\(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
=\(\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=-1\)
Nếu \(a+b+c\ne0\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\Rightarrow\frac{b}{a}=\frac{c}{b}=\frac{a}{c}=1\)
\(P=2.2.2=8\)
Vậy...
Cho \(a^3+b^3+c^3=3abc\)và \(abc\ne0;a+b+c=0\)
CMR \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)=0\)
cho các số hữu tỉ a,b,c thỏa mãn\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)và\(a+b+c\ne0\)tính gt bt\(P=\left(1+\frac{2a}{b}\right)\left(1+\frac{2b}{c}\right)\left(1+\frac{2c}{a}\right)\)
Ta có \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)(dãy tỉ số bằng nhau)
=> a = b = c
Khi đó \(P=\left(1+\frac{2a}{b}\right)\left(1+\frac{2b}{c}\right)\left(1+\frac{2c}{a}\right)=\left(1+\frac{2b}{b}\right)\left(1+\frac{2c}{c}\right)\left(1+\frac{2a}{a}\right)\)
= (1 + 2)(1 + 2)(1 + 2) = 3.3.3 = 27
Vậy P = 27
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\) ( do a + b + c khác 0 )
\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{a}=1\end{cases}}\Rightarrow a=b=c\)
Thế vào P ta được :
\(P=\left(1+\frac{2b}{b}\right)\left(1+\frac{2c}{c}\right)\left(1+\frac{2a}{a}\right)=\left(1+2\right)\left(1+2\right)\left(1+2\right)=27\)
áp dụng dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\) , suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
Nhân vế trên cho 2, ta suy ra: \(\frac{2a}{b}=\frac{2b}{c}=\frac{2c}{a}=2.1=2\)
Thay từng giá trị vào biểu thức P, ta có:
\(P=\left(1+\frac{2a}{b}\right)\left(1+\frac{2b}{c}\right)\left(1+\frac{2c}{a}\right)=\left(1+2\right)\left(1+2\right)\left(1+2\right)=27\)
Vậy giá trị P=27
\(Cho:\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
\(and........a\ne b\ne c........a,b,c\ne0\)
Tính \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
1) Cho \(\frac{a-\left(c-b\right)}{b-c}+\frac{b-\left(a-c\right)}{c-a}+\frac{c-\left(b-a\right)}{a-b}=3\)
CM \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
2) Cho \(\frac{1}{a}+\frac{1}{c}=\frac{1}{b-c}-\frac{1}{a-b}\)và \(ac\ne0\); \(a\ne b\); \(b\ne c\)
CM \(\frac{a}{c}=\frac{a-c}{b-c}\)
1,Cho các số thực a,b,c thỏa mãn điều kiện : \(a^2+b^2+c^2=3\) và \(a+b+c+ab+ac+bc=6\).
Tính \(A=\frac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2014}}\)
2, Cho \(a,b,c\ne0\) thỏa mãn \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=8\),
Chứng minh : \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{3}{4}+\frac{ab}{\left(a+b\right)\left(b+c\right)}+\frac{bc}{\left(b+c\right)\left(c+a\right)}+\frac{ca}{\left(c+a\right)\left(a+b\right)}\)
HELP ME....MAI MÌNH NỘP RỒI
mình cảm ơn
Cho \(a,b,c\in Z;abc\ne0;\frac{a^2+b^2}{2}=ab;\frac{b^2+c^2}{2}=bc;\frac{a^2+c^2}{2}=ac\)
Tính : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)
Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)
\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)
Áp dụng t/c bắc cầu ta dc : \(a=b=c\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)
=>a2+b2=2ab
=>a2-2ab+b2=0
=>(a-b)2=0=>a=b
tương tự=>b=c
=>a=b=c
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)