cho 3 số thực x,y,z khác 0 thỏa mãn: x3 - 27y3 - 8z3 = 18xyz
tính giá trị của biểu thức: P= \(\dfrac{\left(x-3y\right)\left(3xy+2z\right)\left(x-2z\right)}{6xyz}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 12 2020
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
20 tháng 11 2021
Xét \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)
\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)
Xét \(x+y+z\ne0\) thì ta có:
\(\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)
\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)
Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)
20 tháng 11 2021
Nếu bị lỗi thì bạn có thể xem đây nhé:
28 tháng 11 2016
Ta có
\(\frac{x^3}{\left(y+z\right)\left(y+2z\right)}+\frac{y+z}{12}+\frac{y+2z}{18}\ge\frac{3x}{6}=\frac{x}{2}\)
\(\Leftrightarrow\frac{x^3}{\left(y+z\right)\left(y+2z\right)}\ge-\frac{y+z}{12}-\frac{y+2z}{18}+\frac{x}{2}=\frac{18x-7z-5y}{36}\)
Tương tự ta có
\(\frac{y^3}{\left(z+x\right)\left(z+2x\right)}\ge\frac{18y-7x-5z}{36}\)
\(\frac{z^3}{\left(x+y\right)\left(x+2y\right)}\ge\frac{18z-7y-5x}{36}\)
Cộng vế theo vế ta được
\(A\ge\frac{18x-7z-5y}{36}+\frac{18y-7x-5z}{36}+\frac{18z-7y-5x}{36}\)
\(=\frac{x+y+z}{6}\ge\frac{3\sqrt[3]{xyz}}{6}=\frac{3.2}{6}=1\)
Dấu = xảy ra khi x = y = z = 2
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1 2021
1.
Gọi \(d=ƯC\left(2n^2+3n+1;3n+1\right)\)
\(\Rightarrow2n^2+3n+1-\left(3n+1\right)⋮d\)
\(\Rightarrow2n^2⋮d\Rightarrow2n\left(3n+1\right)-3.2n^2⋮d\)
\(\Rightarrow2n⋮d\Rightarrow2\left(3n+1\right)-3.2n⋮d\Rightarrow2⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=2\end{matrix}\right.\)
\(d=2\Rightarrow3n+1=2k\Rightarrow n=2m+1\)
\(\Rightarrow n\) lẻ thì A không tối giản
\(\Rightarrow n\) chẵn thì A tối giản
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1 2021
2.
Giả thiết tương đương:
\(xy^2+\dfrac{x^2}{z}+\dfrac{y}{z^2}=3\)
Đặt \(\left(x;y;\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a^2c+b^2a+c^2b=3\)
Ta có: \(9=\left(a^2c+b^2a+c^2b\right)^2\le\left(a^4+b^4+c^4\right)\left(c^2+a^2+b^2\right)\)
\(\Rightarrow9\le\left(a^4+b^4+c^4\right)\sqrt{3\left(a^4+b^4+c^4\right)}\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)^3\ge81\Rightarrow a^4+b^4+c^4\ge3\)
\(\Rightarrow M=\dfrac{1}{a^4+b^4+c^4}\le\dfrac{1}{3}\)
\(M_{max}=\dfrac{1}{3}\) khi \(\left(a;b;c\right)=\left(1;1;1\right)\) hay \(\left(x;y;z\right)=\left(1;1;1\right)\)
Đặt \(x^3=a^3;27y^3=b^3;8z^3=c^3\)
\(\Rightarrow a^3-b^3-c^3=3abc\)
\(\Rightarrow a^3-b^3-c^3-3abc=0\)
\(\Rightarrow a^3-\left(b+c\right)^3+3bc\left(b+c\right)-3abc=0\)
\(\Rightarrow\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2\right]-3bc\left(a-b-c\right)=0\)
\(\Rightarrow\left(a-b-c\right)\left(a^2+ab+ac+b^2+2bc+c^2-3bc\right)=0\)
\(\Rightarrow\left(a-b-c\right)\left(a^2+b^2+c^2+ab-bc+ca\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a^2+b^2+c^2+ab-bc+ca=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c+a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a=-b=-c\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3y-2z=0\\x=-3y=-2z\end{matrix}\right.\)
*\(x-3y-2z=0\) :
\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}=\dfrac{2z.x.3y}{6xyz}=1\)
*\(x=-3y=-2z\) :
\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}\dfrac{\left(x+x\right)\left(3y+3y\right)\left(-2z-2z\right)}{6xyz}=\dfrac{2x.6y.\left(-4\right)z}{6xyz}=-8\)
Mk sửa lại biểu thức P :\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}\)
Ta có : x3 - 27y3 - 8z3 = 18xyz
<=> (x - 3y)3 + 9xy(x - 3y) - 8z3 = 18xyz
<=> [(x - 3y)3 - (2z)3] + 9xy(x - 3y - 2z) = 0
<=> (x - 3y - 2z)[(x - 3y)2 + (x - 3y).2z + 4z2] + 9xy(x - 3y - 2z) = 0
<=> (x - 3y - 2z)[(x - 3y)2 + (x - 3y).2z + 4z2 + 9zy] = 0
<=> \(\left(x-3y-2z\right)\left\{\left[\dfrac{1}{4}\left(x-3y\right)^2+\left(x-3y\right).2z+4z^2\right]+\dfrac{3}{4}\left(x-3y\right)^2+9xy\right\}=0\)
<=> \(\left(x-3y-2z\right)\left\{\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2\right\}=0\)
<=> \(\left[{}\begin{matrix}x-3y-2z=0\\\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2=0\end{matrix}\right.\)
THI1 x - 3y - 2z = 0
<=> x = 3y + 2z
Khi đó \(P=\dfrac{2z.x.3y}{6xyz}=1\)
TH2 \(\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}\left(x-3y\right)+2z=0\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2z=3y\\x=-3y\end{matrix}\right.\Leftrightarrow x=-3y=-2z\)
Khi đó P = \(\dfrac{\left(-6y\right).\left(-2x\right).\left(-4z\right)}{xyz}=-48\)