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\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
Lấy (2) cộng (3) ta được
\(x^2+y^2-yz-zx=2\) (4)
Lấy (1) - (4) ta được
\(2x\left(x+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-z\end{matrix}\right.\)
Xét 2 TH rồi thay vào tìm được y và z
1. \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{6}{5}\\\dfrac{y+z}{yz}=\dfrac{3}{2}\\\dfrac{z+x}{zx}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{5}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{7}{10}\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
hpt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{1}{2}\\\dfrac{y+z}{yz}=\dfrac{1}{4}\\\dfrac{z+x}{xz}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\) ( đk : x , y , z # 0 )
Cộng từng vế của các pt lại với nhau , ta có :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{12}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{13}{24}-\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{24}-\dfrac{1}{4}=\dfrac{7}{24}\)
\(\Leftrightarrow x=\dfrac{24}{7}\left(tm\right)\)
\(\Rightarrow y=\dfrac{24}{5}\left(tm\right);z=8\left(tm\right)\)
\(\left\{\begin{matrix}x+xy+y=1\\y+yz+z=3\\z+zx+x=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\left(x+1\right)\left(y+1\right)=2\left(1\right)\\\left(y+1\right)\left(z+1\right)=4\left(2\right)\\\left(z+1\right)\left(x+1\right)=8\left(3\right)\end{matrix}\right.\)
Lấy 2(1) - (2) ta được
\(2\left(x+1\right)\left(y+1\right)-\left(y+1\right)\left(z+1\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(2x-z+1\right)=0\)
\(\Leftrightarrow\left\{\begin{matrix}y=-1\\z=2x+1\end{matrix}\right.\)
Với y = -1 thì hệ vô nghiệm
Với z = 2x + 1 thì thế vô 3 được
\(\left(x+1\right)^2=4\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Với x = 1 thì
\(\Rightarrow\left\{\begin{matrix}y=0\\z=3\end{matrix}\right.\)
Với x = - 3 thì
\(\Rightarrow\left\{\begin{matrix}y=-2\\z=-5\end{matrix}\right.\)
\(\left\{\begin{matrix}x+xy+y=1\left(1\right)\\y+yz+z=3\left(2\right)\\z+zx+x=7\left(3\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\left(y+1\right)+\left(y+1\right)=2\\y\left(z+1\right)+\left(z+1\right)=4\\z\left(x+1\right)+\left(x+1\right)=8\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}\left(y+1\right)\left(x+1\right)=2\left(1\right)\\\left(z+1\right)\left(y+1\right)=4\left(2\right)\\\left(x+1\right)\left(z+1\right)=8\left(3\right)\end{matrix}\right.\)(II)
Nhân theo vế: \(\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=2.4.8=64\)
\(\Leftrightarrow\left[\begin{matrix}\left(x+1\right)\left(y+1\right)\left(z+1\right)=-8\left(5\right)\\\left(x+1\right)\left(y+1\right)\left(z+1\right)=8\left(6\right)\end{matrix}\right.\)
(5) và (II) \(\Leftrightarrow\left\{\begin{matrix}z+1=-4\\x+1=-2\\y+1=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}z=-5\\x=-1\\y=-2\end{matrix}\right.\)
(6)và(II)\(\Leftrightarrow\left\{\begin{matrix}z+1=4\\x+1=2\\y+1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}z=3\\x=1\\y=0\end{matrix}\right.\)
Ta có: x2 + y2 + z2 = xy + yz + zx
<=> [(x - y)2 + (y - z)2 + (z - x)2] . 1/2 = 0
<=> x = y = z
Thay vào pt thứ 2...
Natsu Dragneel 2005 pha gần cuối phải là:
\(3.x^{2015}=3.3^{2015}\Leftrightarrow x^{2015}=3^{2015}\Rightarrow x=3\)
ms đúng nha!
AD BĐT cô - si cho ba số không âm x2 ; y2 ; z2 , ta có :
x2 + y2 ≥ 2√x2y2 = 2xy ( dấu bằng xảy ra khi x = y )
Tương tự : y2 + z2 ≥ 2yz ( dấu ... khi y = x )
z2 + x2 ≥ 2zx ( ... z = x )
⇒ 2 ( x2 + y2 + z2 ) ≥ 2 ( xy + yz + zx )
⇔ x2 + y2 + z2 ≥ xy + yz + zx
Dấu = xảy ra khi x = y = z
⇒ x2015 + y2015 + z2015 = 3x2015 = 32016
⇔ 32015. x = 32015. 3 ⇒ x = 3
⇒ x = y = z = 3
\(\left\{{}\begin{matrix}xy=x+y+1\\yz=y+z+5\\zx=z+x+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-x-y-1=-2\\yz-y-z-1=4\\zx-z-x-1=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=-2\\\left(y-1\right)\left(z-1\right)=4\\\left(z-1\right)\left(x-1\right)=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y+z=3\left(2\right)\\x+2y-z=2\left(1\right)\\x+yz+zx\left(2\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3y=5\\2z-y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3y=5\\10z-5y=5\end{matrix}\right.\Rightarrow2x+3y-10z+5y=0\Rightarrow2x+8y=10z\Rightarrow x+4y=5z\)tối tớ về lm tiếp