Tìm x, y, z biết
\(x^2+y^2+z^2=x\left(y+z\right)\)
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PL
28 tháng 7 2017
Ta có:
\(x\left(x+y+z\right)=\frac{15}{2}\)
\(y\left(x+y+z\right)=\frac{-5}{2}\)
\(z\left(x+y+z\right)=20\)
=>\(x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=\frac{15}{2}+\frac{-5}{2}+20\)
\(\left(x+y+z\right)\left(x+y+z\right)=\frac{15-5}{2}+20\)
\(\left(x+y+z\right)^2=\frac{10}{2}+20\)
\(\left(x+y+z\right)^2=5+20\)
\(\left(x+y+z\right)^2=25\)
=>x+y+z=5 hoặc x+y+x=-5
Với x+y+z=5
=>\(x.5=\frac{15}{2}\)=>\(x=\frac{15}{2}.\frac{1}{5}=\frac{3}{2}\)
\(y.5=\frac{-5}{2}\)=>\(y=\frac{-5}{2}.\frac{1}{5}=\frac{-1}{2}\)
\(z.5=20\)=>\(z=\frac{20}{5}=4\)
Với x+y+z=-5
=>\(x.\left(-5\right)=\frac{15}{2}\)=>\(x=\frac{15}{2}.\frac{-1}{5}=\frac{-3}{2}\)
\(y.\left(-5\right)=\frac{-5}{2}\)=>\(y=\frac{-5}{2}.\frac{-1}{5}=\frac{1}{2}\)
\(z.\left(-5\right)=20\)=>\(z=\frac{20}{-5}=-4\)
Vậy \(x=\frac{3}{2},y=-\frac{1}{2},z=4\); \(x=-\frac{3}{2},y=\frac{1}{2},z=-4\)
28 tháng 7 2017
Ta có:
\(x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=\frac{15}{2}+\left(-\frac{5}{2}\right)+20\)(Cộng vế với vế)
\(\Leftrightarrow\left(x+y+z\right)\left(x+y+z\right)=\frac{50}{2}=25\)
\(\Rightarrow\left(x+y+z\right)^2=25\Leftrightarrow x+y+z=\sqrt{25}=5\)
\(\Rightarrow\hept{\begin{cases}x.5=\frac{15}{2}\Rightarrow x=\frac{3}{2}\\y.5=-\frac{5}{2}\Rightarrow y=-\frac{1}{2}\\z.5=20\Rightarrow z=4\end{cases}}\)
Vậy \(x=\frac{3}{2};y=-\frac{1}{2};z=4\).
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 2 2022
Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)
\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)
21 tháng 11 2021
Gọi cái biểu thức đó là P nha
Trước tiên chứng minh:
\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\left(\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right)=0\)
\(\Leftrightarrow\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\Leftrightarrow x-y+y-z+z-x=0\)( đúng )
Giờ ta quay lại bài toán ban đầu
Ta có:
\(\Leftrightarrow2P=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\ge\frac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\frac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}+\frac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)
\(\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)}+\frac{\left(y+z\right)^2}{4\left(y+z\right)}+\frac{\left(z+x\right)^2}{4\left(z+x\right)}\)
\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}=\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{4}\)
LT
2
18 tháng 3 2017
Ta có : \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
mà \(\hept{\begin{cases}\left(x-y^2+z\right)^2\ge0\\\left(y-2\right)^2\ge0\\\left(z+3\right)^2\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-y^2+z=0\\y-2=0\\z+3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=y^2-z=2^2-\left(-3\right)=7\\y=2\\z=-3\end{cases}}\)
Q
18 tháng 3 2017
\(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)
Do \(\hept{\begin{cases}\left(x-y^2+z\right)^2\ge0\\\left(y-2\right)^2\ge0\\\left(z+3\right)^2\ge0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\y=2\\z=-3\end{cases}\Leftrightarrow\hept{\begin{cases}\left[x-2^2+\left(-3\right)\right]^2=0\\y=2\\z=-3\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\y=2\\z=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\\z=-3\end{cases}}}\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 12 2020
\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(P=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}=\dfrac{1}{3}\)
MP
1
24 tháng 12 2021
\(\Rightarrow\dfrac{z+y+1}{x}=\dfrac{x+z+1}{y}=\dfrac{x+y-2}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2=x+y+z\\ \Rightarrow\left\{{}\begin{matrix}z+y+1=2x\\x+z+1=2y\\x+y-2=2z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x-1\\x+z=2y-1\\x+y=2z+2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x-1=2-x\\2y-1=2-y\\2z+2=2-z\end{matrix}\right.\Rightarrow\left(x,y,z\right)=\left(1;1;0\right)\)
Ta có: \(x^2+y^2+z^2=x\left(y+z\right)\)
<=> \(2x^2+2y^2+2z^2=2x\left(y+z\right)\)
<=> \(\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+y^2+z^2=0\)
<=> \(\left(x-y\right)^2+\left(x-z\right)^2+y^2+z^2=0\)(1)
Lại có: \(\left(x-y\right)^2\ge0;\left(x-z\right)^2\ge0;y^2\ge0;z^2\ge0\) với mọi x; y; z
Ta có: \(\left(x-y\right)^2+\left(x-z\right)^2+y^2+z^2\ge0\) với mọi x; y; z
Nên (1) <=> x - y = x - z = y = z = 0 <=> x = y = z = 0.