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1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)
\(\Rightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=45\)
\(\Rightarrow x^2+y^2-2xy=9\)
\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)
\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)
Vậy...
1) Đặt \(B=x^2+y^2+z^2\)
\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)
\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)
Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)
Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
Sửa đề :
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\)
Bài làm
đề có sai chỗ nào ko bn,mk thấy chỗ giả thiết sai sai thì phải,bn kt lại giúp mk
Bài 1
\(a^2-2a+6b+b^2=-10\)
<=>\(a^2-2a+1+b^2+6b+9=0\)
<=>\((a-1)^2+(b+3)^2=0\)
Ta lại có: \((a-1)^2\ge0 \)
\((b+3)^2\ge0\)
=> \((a-1)^2+(b+3)^2\ge0\)
Mà\((a-1)^2+(b+3)^2=0\)
=>(a-1)2=0=>a=1
(b+3)2=0=>b=-3
Vậy a=1,b=-3
Bài 2
Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}= \frac{x+y}{z}+1+\frac{x+z}{y}+1+ \frac{y+z}{x}+1 -3 \)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3=(x+y+z)( \frac{1}{z}+\frac{1}{x}+\frac{1}{y})-3=0-3=-3 \)
Ta có : \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Leftrightarrow2axby+2axvz+2bycz=a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-2axby-2azcx-2bycz=0\)
\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Do \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
:D
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
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\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)