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Cách 2:
\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)
\(\Rightarrow\dfrac{30x^2}{60}+\dfrac{20y^2}{60}+\dfrac{15z^2}{60}=\dfrac{12\left(x^2+y^2+z^2\right)}{60}\)
\(\Rightarrow\dfrac{30x^2+20y^2+15z^2}{60}=\dfrac{12x^2+12y^2+12z^2}{60}\)
\(\Rightarrow30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\)
\(\Rightarrow30x^2-12x^2+20y^2-12y^2+15z^2-12z^2=0\)
\(\Rightarrow18x^2+8y^2+3z^2=0\)
Ta có :
\(18x^2\ge0\forall x\) \(;8y^2\ge0\forall y;3z^2\ge0\forall z\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}18x^2=0\\8y^2=0\\3z^2=0\end{matrix}\right.\Leftrightarrow x=y=z=0\)
Vậy x = y = z =0
Ta có : \(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)
\(\Leftrightarrow\dfrac{30x^2+20y^2+15z^2}{60}=\dfrac{12\left(x^2+y^2+z^2\right)}{60}\)
\(\Leftrightarrow30x^2-18x^2+20y^2-12y^2+15z^2-12z^2=0\)
\(\Leftrightarrow18x^2+8y^2+3z^2=0\)
Vì \(\left\{{}\begin{matrix}18x^2\ge0\\8y^2\ge0\\3z^2\ge0\end{matrix}\right.\Rightarrow18x^2+8y^2+3z^2\ge0\)
Dấu '' = '' xảy ra \(\Leftrightarrow x=y=z=0\)
Vậy với \(x=y=z=0\) thì \(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\)
1) a) \(\dfrac{x^2-y^2}{x^3}+y^{^3}.\left(\dfrac{xy-x^2-y^2}{y}.\dfrac{xy}{y-x}\right)\)
\(=\dfrac{x^2-y^2}{x^3}+y^3.\dfrac{x\left(xy-x^2-y^2\right)}{y-x}\)
\(=\dfrac{x^2-y^2}{x^3}+\dfrac{xy^3\left(xy-x^2-y^2\right)}{y-x}\)
\(=\dfrac{-\left(x-y\right)^2\left(x+y\right)+xy^3\left(xy-x^2-y^2\right)}{x^3\left(y-x\right)}\)
Cậu tự thu gọn nốt nhé , tớ sắp đi hok
Bài 2 . Theo giả thiết : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
=> \(\dfrac{yz+xz+xy}{xyz}=\dfrac{1}{x+y+z}\)
=> \(\left(x+y+z\right)\left(yz+zx+xy\right)=xyz\)
=>\(x\left(yz+xz+xy\right)+y\left(yz+xz+xy\right)+z\left(yz+xz+xy\right)-xyz=0\)=> \(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
Ta có :
* x = - y
* y = -z
* x = -z
Áp dụng đều này vào phân thức cần CM , ta có :
TH1 . x = -y
\(\dfrac{1}{\left(-y\right)^5}+\dfrac{1}{y^5}+\dfrac{1}{z^5}=\dfrac{1}{\left(-y\right)^5+y^5+z^5}\)
=> \(\dfrac{1}{z^5}=\dfrac{1}{z^5}\), luôn đúng
Tương tự thử với các trường hợp còn lại ta cũng sẽ có được đpcm
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
B1:
pt <=> \(\dfrac{3x^2}{10}+\dfrac{2y^2}{15}+\dfrac{z^2}{20}=0\)
<=> x = y = z = 0
B2: Áp dụng bđt Cô-si:
\(\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)\ge2+2=4\)
Dấu "=" xảy ra <=> x2 = y2 = 1
a/\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2y}=\dfrac{54}{2y}\)
\(\Rightarrow2y\cdot y=54\cdot3\Rightarrow2y^2=162\Rightarrow y^2=\dfrac{162}{2}=81\)
Mà y > 0 (gt) => \(y=\sqrt{81}=9\Rightarrow x=\dfrac{54}{9}=6\)
Vậy..............
b/ \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{4}\cdot25=\dfrac{25}{4}\\y^2=\dfrac{1}{4}\cdot9=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\sqrt{\dfrac{25}{4}}=\pm\dfrac{5}{2}\\y=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\end{matrix}\right.\)
Vậy.............
c/ x/2 = y/3 => x/10 = y/15
y/5 = z/7 => y/15 = z/21
=> x/10 = y/15 = z/21
Áp dụng t/c của dãy tỉ số = nhau là ra....
1) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+2.x.3+3^2}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x+3}{x-3}=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy x=-3
bạn ơi x ko thể bằng -3 đc vì
\(\dfrac{x}{x+3}=\dfrac{-3}{-3+3}=\dfrac{-3}{0}\) là sai
a, \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{x^2+y^2}{4+16}=\dfrac{2000}{20}=100\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=100.4=400\\y^2=100.16=1600\\z^2=100.25=2500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm20\\y=\pm40\\z=\pm50\end{matrix}\right.\)
Do \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}x=20\\y=40\\z=50\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-20\\y=-40\\z=-50\end{matrix}\right.\)
Vậy ...
b, \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{14-6}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.3=3\\z-3=1.4=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
Vậy ...
c, \(x-z=-2\Rightarrow x+2=z\)
Do đó \(y.z=12\Leftrightarrow y.\left(x+2\right)=12\Rightarrow xy+2y=12\Rightarrow6+2y=12\)
\(\Rightarrow y=3\Rightarrow x.3=6\Rightarrow x=2\Rightarrow2-z=-2\Rightarrow z=4\)
Vậy x=2; y=3; z=4
\(\dfrac{x^2.30+y^2.20+z^2.15}{60}=\dfrac{12\left(x^2+y^2+z^2\right)}{60}\\ \Rightarrow30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\\ \Rightarrow30x^2-12x^2+20y^2-12y^2+15z^2-12z^2=0\\ \Rightarrow18x^2+8y^2+3z^2=0\\ \)
Mà \(x^2\ge0\forall x\\ y^2\ge0\forall y\\ z^2\ge0\forall z\)
=> \(\left\{{}\begin{matrix}18x^2=0\\8y^2=0\\3z^2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)
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