1, Tính giá trị biểu thức:
P=(1+a/b) (1+b/c) (1+c/a)
2, Cho 1/a+1/b+1/c=0
Tính A=bc/a^2+ca/b^2+ab/c^2
3,Cho x+y+z=0
Chứng minh rằng 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)
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DA
20 tháng 10 2018
Bài 1: theo mình nghĩ thì nên cho thêm điều kiện gì chứ ạ :(
Bài 2: Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{ab}.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\) ( hằng đẳng thức: \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\) )
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{ab}.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{ab}.\left(-\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Có \(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)
\(A=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
\(A=abc.\dfrac{3}{abc}=3\)
Bải 3: Ta có
\(x+y+z=0\)
\(\Rightarrow y+z=-x\)
\(\Rightarrow\left(y+z\right)^5=-x^5\)
\(\Rightarrow y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left[\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right]=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2-yz+z^2+2yz\right)=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(\Rightarrow x^5+y^5+z^5=-5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(\Rightarrow2\left(x^5+y^5+z^5\right)=2.-5yz.\left(-x\right)\left(y^2+yz+z^2\right)\)
\(\Rightarrow2.\left(x^5+y^5+z^5\right)=5xyz.\left(2y^2+2yz+2z^2\right)\)
\(\Rightarrow2\left(x^5+y^5+z^5\right)=5xyz\left[\left(y+z\right)^2+y^2+z^2\right]\)
\(\Rightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
MY
1
10 tháng 5 2021
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\) \(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\) \(\Rightarrow ayz+bxz+cxy=0\) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\) \(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)