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![](https://rs.olm.vn/images/avt/0.png?1311)
`A + B + C = x^2yz + xy^2z + zy^2x = xyz(x+y+z) = xyz`.
\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)
Vậy ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có A+B+C=x2yz+xy2z+xyz2
=x(xyz)+y(xyz)+z(xyz)
=x.1+y.1+z.1
=x+y+z(dpcm)
\(A=x^2yz=x.\left(xyz\right)=x.1=x\)
\(B=xy^2z=y.\left(xyz\right)=y.1=y\)
\(C=xyz^2=z.\left(xyz\right)=z.1=z\)
\(\Rightarrow A+B+C=x+y+z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chứng tỏ rằng đa thức \(x^{2008}-x^{2007}+1\) vô nghiệm hay gì vậy ạ :v?
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)
c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề bài ta có x = \(\frac{a}{m}\) , y = \(\frac{b}{m}\)( a, b, m \(\in\) Z, m > 0 )
Vì x < y nên ta suy ra a < b
Ta có : x = \(\frac{2a}{2m}\), y = \(\frac{2b}{2m}\), , z = \(\frac{a+b}{2m}\)
Vì a < b => a + a < a +b => 2a < a + b
Do 2a< a +b nên x < z (1)
Vì a < b => a + b < b + b => a + b < 2b
Do a+b < 2b nên z < y (2)
Từ (1) và (2) ta suy ra x < z< y
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: a+b+c=1
<=>(a+b+c)^2=1
<=>ab+bc+ca=0 (1)
mặt khác: áp dụng tính chất dãy tỉ số bằng nhau ta có:
x/a=y/b=z/c=(x+y+z)/(a+b+c)=x+y+z
<=> x=a(x+y+z) ; y=b(x+y+z) ; z=c(x+y+z)
=>xy+yz+zx=ab(x+y+z)^2+bc(x+y+z)^2+ca(x...
<=>xy+yz+zx=(ab+bc+ca)(x+y+z)^2 (2)
từ (1) và (2) ta có đpcm
A+B+C=\(X^2\)YZ+X\(Y^2\)Z+XY\(Z^2\)=XXYZ+XYYZ+XYZZ=(X+Y+Z)XYZ
MÀ XYZ=1=>A+B+C=(X+Y+Z)*1=X+Y+Z
cm: A+B+C = xyz cơ mà