Consider three positive integers so that: \(\frac{3}{x-1}=\frac{4}{x-2}=\frac{5}{x-3}\) and \(xyz=192\). Find x, y, z.
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The rectangle ABCD is divided into 4 regions whose perimeters are indicated in the figure below,where X,Y,Z Are Distinct positive integers and X>Y .It is known that Z=\(\frac{X+Y}{3}\)and W<6.Find X
The rectangle ABCD is divided into 4 regions whose perimeters are indicated in the figure below,where X,Y,Z are distinct positive integers and X>Y .It is known that Z=\(\frac{Z+Y}{3}\)and W<6.Find X
20 tháng 4 2021 lúc 23:12
Let x, y, z be positive real numbers such that xy + yz + zx + xyz = 4 . Prove that :
\(3\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\ge\left(x+2\right)\left(y+2\right)\left(z+2\right)\)
Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh
\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng holder ta có:
\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)
\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)
\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)
Từ đây ta cần chứng minh:
\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )
Vậy có ĐPCM
Tìm x,y,z biết
1. \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)và xyz=-30
2.\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)và \(x^2+y^2-z^2\)=-12
3.\(\frac{x}{3}=\frac{y}{2}=\frac{z}{4}\)và xyz=192
Tìm x,y,z biết
\(\frac{3}{x-1}=\frac{4}{y-2}=\frac{5}{z-3}\) và xyz=192
Cho \(\frac{3}{x-1}\)=\(\frac{4}{y-2}\)=\(\frac{5}{z-3}\)và xyz=192. Tìm x,y,z
ta có ........= x-1/3 ........ ( đảo ngược lại) =k
=> x-1 =3k=> x = 3k+1
y-2 = 4k=> y =4k+2
z-3=5k=> z = 5k+3
=> xyz = 3k+1. 4k+2.5k+3= 192 . (1+2+3) =.....
=> tự tính nha bn
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Đặt \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}=k\) ( đảo ngược )
\(\Rightarrow x-1=3k;y-2=4k;z-3=5k\)
\(\Rightarrow x=3k+1;y=4k+2;z=5k+3\)
\(\Rightarrow xyz=192\Leftrightarrow\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)=192\Leftrightarrow k=1\)
Suy ra :
+) \(x-1=3k\Leftrightarrow x-1=3\Leftrightarrow x=4\)
+) \(y-2=4k\Leftrightarrow y-2=4\Leftrightarrow y=6\)
+) \(z-3=5k\Leftrightarrow z-3=5\Leftrightarrow z=8\)
Vậy x = 4 ; y = 6 ; z = 8
Tìm x,y,z biết
\(\frac{3}{x-1}\)=\(\frac{4}{y-2}\)=\(\frac{5}{z-3}\) và xyz =192
Suppose that x, y, z are positive integers such that x > y > z > 663 and x, y, z satisfy x + y + z = 1998 and 2x + 3y + 4z = 5992. Find x, y, z
thằng này số nhọ , hai năm rồi méo có ai trả lời
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let x,y,z>0 such that xyz=1. show that \(\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{x^4+x+y}}\ge2\sqrt{xy+yz+zx}\)
