cho x+1/x=a. Tính x^7+1/x^7 theo a
Cho x, y, z khác 0 thỏa mãn đồng thời 1/x +1/y + 1/z =2 và 2/xy - 1/z^2 = 4 Tính giá trị của biểu thức p=(x+2y+z)^2018
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
24 tháng 5 2019
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)
Khai triển cả 2 vế ta được \(\left(\frac{1}{y}+\frac{1}{z}\right)^2+\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)
=>\(\hept{\begin{cases}\frac{1}{y}+\frac{1}{z}=0\\\frac{1}{x}+\frac{1}{z}=0\end{cases}}\)=>\(\frac{1}{x}=\frac{1}{y}\Rightarrow x=y\)
=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{x}+\frac{1}{z}=2\Rightarrow\frac{4}{x^2}+\frac{4}{xz}+\frac{1}{z^2}=4\)(1)
\(\frac{2}{xy}-\frac{1}{z^2}=\frac{2}{x^2}-\frac{1}{z^2}=4\)(2)
Từ (1) và (2) suy ra
\(\frac{2}{x^2}+\frac{4}{xz}+\frac{2}{z^2}=0\Rightarrow\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}=0\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)\(\Rightarrow\frac{1}{x}+\frac{1}{z}=0\Rightarrow x=y=-z\)
=> \(P=\left(x+2y+z\right)^{2019}=\left(2y\right)^{2019}\)
24 tháng 5 2019
à thêm cái này nữa. Sorry viết thiếu
Vì x=y=-z\(\Rightarrow\frac{2}{x}-\frac{1}{x}=2\Rightarrow\frac{1}{x}=2\Rightarrow x=\frac{1}{2}.\)
lúc đó \(P=\left(2.\frac{1}{2}\right)^{2019}=1\)
7 tháng 6 2018
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2013}=\frac{1}{x+y+z}\Rightarrow\frac{yz+xz+xy}{xyz}=\frac{1}{x+y+z}\Rightarrow\left(yz+xz+xy\right)\left(x+y+z\right)=xyz\)
\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz+xyz=xyz\)
\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz=0\)
\(\Rightarrow\left(x^2y+x^2z+xy^2+xyz\right)+\left(y^2z+xz^2+y^2z+xyz\right)=0\)
\(\Rightarrow x\left(xy+xz+y^2+yz\right)+z\left(yz+xz+y^2+xy\right)=0\)
\(\Rightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+y=0\Rightarrow x^3+y^3=0\\y+z=0\Rightarrow y^5+z^5=0\\x+z=0\Rightarrow z^7+x^7=0\end{cases}}\)
\(\Rightarrow A=\left(x^3+y^3\right)\left(y^5+z^5\right)\left(z^7+x^7\right)=0\)
PN
5 tháng 9 2020
chắc câu này a đăng lên cho vui :vv
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2< =>\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2=4\)
\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)
\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\left(\frac{2}{xy}-\frac{1}{z^2}\right)+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}+4=4\)
\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{xy}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4-4\)
\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)
\(< =>\left(\frac{1}{x^2}+\frac{2}{zx}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(< =>\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0< =>\frac{1}{x}=\frac{1}{y}=-\frac{1}{z}\)
\(< =>x=y=-z\)Thế vào giả thiết ta được : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(< =>\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2< =>\frac{-1}{z}+\frac{-1}{z}+\frac{1}{z}=2\)
\(< =>\frac{-1-1+1}{z}=2< =>2z=-1< =>z=-\frac{1}{2}\)
Suy ra \(x=y=-z=-\left(-\frac{1}{2}\right)=\frac{1}{2}< =>\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{cases}}\)
Nên \(P=\left(x+2y+z\right)^{2019}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2019}=1^{2019}=1\)
Bài 1 :
Ta có :
\(x^7+\frac{1}{x^7}=\left(x^3+\frac{1}{x^3}\right)\left(x^4+\frac{1}{x^4}\right)-\left(x+\frac{1}{x}\right)\)
\(\left(x+\frac{1}{x}\right)=a\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-x.\frac{1}{x}+\frac{1}{x^2}\right)\)
\(=a\left(x^2+\frac{1}{x^2}-1\right)=a\left(a^2-3\right)\)
\(x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2.x^2.\frac{1}{x^2}\)
\(=\left(a^2-2\right)^2-2=a^4-4a^2+4-2\)
\(=a^4-4a^2+2\)
\(\Rightarrow x^7+\frac{1}{x^7}=a.\left(a^2-3\right).\left(a^4-4a^2+2\right)-a\)
\(=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a\)
\(=a^7-4a^5+2a^3-3a^5+12a^3-6a-a\)
\(=a^7-7a^5+14a^3-7a\)
Bài 2 :
Ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)
\(\Rightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{z}=\frac{1}{y}+\frac{1}{z}=0\) vì \(\left(\frac{1}{x}+\frac{1}{z}\right)^2,\left(\frac{1}{y}+\frac{1}{z}\right)^2\ge0\)
\(\Rightarrow x=y=-z\)
\(\Rightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\Rightarrow-\frac{1}{z}=2\Rightarrow z=-\frac{1}{2}\)
\(\Rightarrow x=y=\frac{1}{2}\)
\(\Rightarrow x+2y+z=\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}=1\)
\(\Rightarrow P=1\)