Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2 ) b )
\(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)
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
\(x+y+z=0\)
\(\Rightarrow y+z=-x\)
\(\Rightarrow\left(y+z\right)^5=-x^5\)
\(\Rightarrow y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left[\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right]=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2-yz+z^2+2yz\right)=0\)
\(\Rightarrow x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(\Rightarrow x^5+y^5+z^5=-5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(\Rightarrow2\left(x^5+y^5+z^5\right)=2.-5yz.\left(-x\right)\left(y^2+yz+z^2\right)\)
\(\Rightarrow2.\left(x^5+y^5+z^5\right)=5xyz.\left(2y^2+2yz+2z^2\right)\)
\(\Rightarrow2\left(x^5+y^5+z^5\right)=5xyz\left[\left(y+z\right)^2+y^2+z^2\right]\)
\(\Rightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
Lời giải:
Khai triển:
\(\text{VT}=5(x^5+y^5+z^5)+5\underbrace{[x^3(y^2+z^2)+y^3(x^2+z^2)+z^3(x^2+y^2)]}_{M}\)
Xét riêng $M$ kết hợp với điều kiện $x+y+z=0$ ta có
\(M=x^2y^2(x+y)+y^2z^2(y+z)+z^2x^2(x+z)=-(x^2y^2z+y^2z^2x+z^2x^2y)\)
\(\Leftrightarrow M=-xyz(xy+yz+xz)=\frac{-1}{2}xyz[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}xyz(x^2+y^2+z^2)\)
Ta biết đến một hằng thức rất quen thuộc: Nếu $x+y+z=0$ thì \(x^3+y^3+z^3=3xyz\)
Cách chứng minh: \(x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=0-3(-x)(-y)(-z)=3xyz\)
Do đó \(M=\frac{1}{6}(x^3+y^3+z^3)(x^2+y^2+z^2)=\frac{\text{VT}}{30}\)
\(\Rightarrow \text{VT}=5(x^5+y^5+z^5)+5M=5(x^5+y^5+z^5)+\frac{\text{VT}}{6}\)
\(\Rightarrow \text{VT}=6(x^5+y^5+z^5)\) (đpcm)
b) Theo phần a)
\(\left\{\begin{matrix} M=\frac{1}{2}xyz(x^2+y^2+z^2)\\ M=\frac{5(x^2+y^2+z^2)(x^3+y^3+z^3)}{30}\end{matrix}\right.\Rightarrow \frac{5(x^2+y^2+z^2)(x^3+y^3+z^3)}{30}=\frac{xyz(x^2+y^2+z^2)}{2}\)
Mà \(5(x^2+y^2+z^2)(x^3+y^3+z^3)=6(x^5+y^5+z^5)\Rightarrow \frac{6(x^5+y^5+z^5)}{30}=\frac{xyz(x^2+y^2+z^2)}{2}\)
\(\Leftrightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)\) (đpcm)
b)Vì x+y+z=0
=>x+y=-z =>(x+y)^5=-z^5
hay x^5+y^5+5(x^4y+xy^4+2x³y²+2x²y³+)=-z^5
<=>x^5+y^5+z^5+5xy(x³+y³+2x²y+2x²y)=0
<=>x5+y^5+z^5+5xy(x+y)(x²-xy+y²+2xy)=0
<=>x^5+y^5+z^5-5xyz(x²+xy+y²)=0
<=>x^5+y^5+z^5=5xyz(x²+xy+y²)
<=>2(x^5+y^5+z^5)=5xyz(2x²+2xy+2y²)
<=>2(x^5+y^5+z^5)=5xyz[x²+y²+(x+y)²]
<=>2(x^5+y^5+z^5)=5xyz(x³+y²+z²)
