C/m (x+y+z)^3 -x^3- y^3-z^3 = 3(x+y)(y+z)(z+x)
Cho x,y,z>0 C/M
\(\sqrt{\dfrac{x^3}{y^3}}+\sqrt{\dfrac{y^3}{z^3}}+\sqrt{\dfrac{z^3}{x^3}}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\)
\(\sqrt{\dfrac{x^3}{y^3}}+\sqrt{\dfrac{x^3}{y^3}}+1\ge\dfrac{3x}{y}\) ; \(2\sqrt{\dfrac{y^3}{z^3}}+1\ge\dfrac{3y}{z}\) ; \(2\sqrt{\dfrac{z^3}{x^3}}+1\ge\dfrac{3z}{x}\)
\(\Rightarrow2VT+3\ge\dfrac{3x}{y}+\dfrac{3y}{z}+\dfrac{3z}{x}\)
\(\Rightarrow2VT+3\ge\dfrac{2x}{y}+\dfrac{2y}{z}+\dfrac{2z}{x}+3\sqrt[3]{\dfrac{xyz}{xyz}}\)
\(\Rightarrow VT\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\) (đpcm)
c/m vs mọi số nguyên x, y, z thì
P=(x+y+z)^3-(y+z-x)^3-(x+z-y)^3-(x+y-z)^3 chia hết cho 24
Đặt y+z-x=a
x+z-y=b
x+y-z=c
Ta thấy a+b+c=y+z-x+x+z-y+x+y-z=x+y+z
Ta có: \(P=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3-c^3\)
\(=3a^2b+3ab^2+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(=3\cdot2z\cdot2y\cdot2x\)
\(=24xyz⋮24\)
Vậy P chia hết cho 24
Tìm x,y,z biết:
a, x = y/6 = z/3
b, x/2 = y = z/3
c, x/6 = y/3 = z/3
d, x/2 = y/3 = z/4
e, x/2 = y/-2 = z/5
f, x/2 = y/-3 = z/4
cho x;y;z >0 thỏa mãn x+y+z=2008. c/m : \(\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^3+x^3}\ge2008\)
Tìm x,y,z biết:
a) x/15=y/20=z/28 và 2x+2y-z=186
b)x/3=y/4 ; y/5=2x-z=-3y+372
c)3x=2y ; 7y=5z và x+y+z=98
d)3x=2y=4z và x+y+z=104
e)1/2.x=2/3.y=3/4.z và x-y=15
g) x-1/2=y-2/3=z-3/4 và 2x+3y-2=50
h) y+2+1/x=x+z+2/y=x+y-3/z=1/x+y+z
i)x/y+z+1=y/x+z+1=z/x+y-2=x+y+z
k)x/2=y/3=z/5 và x.y.z=810
m) x/y=2/3 ; x/z=4/9 và x^3+y^3+z^3=1009
giúp vs mình đg gấp ạ, tks
a) (x+y)(x^2-y^2)+(y+z)(y^2-z^2)+(z+x)(z^2-x^2)
b) x^3(y-z)+y^3(z-x)+z^3(x-y)
c)x^3(z-y)+y^3(x-z)+z^3(y-z)+xyz(xyz-1)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
cho x,y,z nguyên và (x-y)*(y-z)*(z-x)=m. Chứng minh rằng: (x-y)^3 + (y-z)^3 + (z-x)^3 chia hết cho m
Một bài toán "lừa" người ta:
Đặt \(a=x-y,b=y-z,c=z-x\Rightarrow a+b+c=0\).
Ta có hằng đẳng thức \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).
Trong trường hợp này thì \(a+b+c=0\) nên suy ra đpcm.
cho x;y ;z>0 thỏa mãn x+y+z=2008
c/m \(\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^3+x^3}\ge2008\)