cho x/y=y/z/=z/t .chung minh rang:(x+y+z/y+z+t)^3=x/t
H24
Những câu hỏi liên quan
Chung minh rang:(x-y)+(z-t)=(x+z)-(y+t)
cho x/y+z+t=y/z+t+x=z/t+x+y=t/x+y+z
chung minh bieu thuc sau co gia tri nguyen
p=(x+y/z+t)+(y+z/t+x)+(z+t/x+y)+(t+x/y+z)
Cho x;y;z;t€Z
Chung minh (x-y)(x-z)(x-t)(y-z)(y-t)(z-t) chia het cho 12
Cach lam nua nhe
cho x;y;z;t la 4 so khac 0 va thoa man cac dieu kien sau:
y^2=xz, z^2=yt, vay^3+z^3+t^3kac 0chung minh rang:
(y^3+z^3+x^3)/y^3+z^3+t^3=x/t
Chung minh A= x/y+z+t =y/z+t+x = z/t+x+y = t/x+y+z la so nguyen.
tham khảo: https://hoidap247.com/cau-hoi/1025167
cho x=a/b va y=b/m voi a,b,m thuoc Z ;m>0 va x<y .Chung minh rang Z=a+b/2m thi x<Z<t
Chung minh rang
(x+y+z)^3 = x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)
12
24
36
48
60
72
84
96
208
120
132144
156
168
180
Đúng 0
Bình luận (0)
cho các số dương x,y,z,t . Chứng minh: \(\frac{40}{3}\le\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\)
\(VP=\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}=\left(\frac{x}{y+z+t}+\frac{y+z+t}{9x}\right)+\left(\frac{y}{z+t+x}+\frac{z+t+x}{9y}\right)+\left(\frac{z}{t+x+y}+\frac{t+x+y}{9z}\right)+\left(\frac{t}{x+y+z}+\frac{x+y+z}{9t}\right)+\frac{8}{9}\left(\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\right)\)\(\ge8\sqrt[8]{\frac{x}{y+z+t}.\frac{y}{z+t+x}.\frac{z}{t+x+y}.\frac{t}{x+y+z}.\frac{y+z+t}{9x}.\frac{z+t+x}{9y}.\frac{t+x+y}{9z}.\frac{x+y+z}{9t}}+\frac{8}{9}\left(\frac{y}{x}+\frac{z}{x}+\frac{t}{x}+\frac{z}{y}+\frac{t}{y}+\frac{x}{y}+\frac{t}{z}+\frac{x}{z}+\frac{y}{z}+\frac{x}{t}+\frac{y}{t}+\frac{z}{t}\right)\)\(\ge\frac{8}{3}+\frac{8}{9}.12\sqrt[12]{\frac{y}{x}.\frac{z}{x}.\frac{t}{x}.\frac{z}{y}.\frac{t}{y}.\frac{x}{y}.\frac{t}{z}.\frac{x}{z}.\frac{y}{z}.\frac{x}{t}.\frac{y}{t}.\frac{z}{t}}=\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}=VT\left(đpcm\right)\)
Đẳng thức xảy ra khi x = y = z = t > 0
cho 3 so nguyen x,y,z thoa man x+y+z=0 chung minh rang x^3+y^3+z^3= 3xyz
xét hiệu x3+y3+z3-3xyz
=(x+y)3+z3-3xy(x+y)-3xyz
=(x+y+z)3-3(x+y+z)(x+y)z-3xy(x+y+z)
=0 vì x+y+z=0
=>x3+y3+z3=3xyz
=>đpcm
Đúng 0
Bình luận (0)