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A + B + C = x2.y.z + x.y2.z + x.y.z2 = x.y.z.(x + y + z) = x.y.z .1 = xyz (Vì x+ y + z = 1)
\(x^2=y.z\Rightarrow\frac{x}{y}=\frac{z}{x}\)
tuong tự ta có\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}=\frac{x+z+y}{y+x+z}=1\)
=> dpcm
Lile nhá bạn
Ta có:
\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{z+x}{zx}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\)
Thay tất cả giá trị x,y,z vào M ta được:
\(M=\frac{2020x^3+2020y^3+2020z^3}{x^3+y^3+z^3}+\frac{2021x^5+2021y^5}{x^5+y^5}\)
\(\Rightarrow M=\frac{2020\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}+\frac{2021\left(x^5+y^5\right)}{x^5+y^5}\)
\(\Rightarrow M=2020+2021=4041\)
Lời giải:
Ta có:
\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}\)
\(A=\frac{xz}{xyz+xz+z}+\frac{y.xz}{yz.xz+y.xz+xz}+\frac{z}{zx+z+1}\)
\(A=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\) (thay \(xyz=1\) )
\(A=\frac{xz+1+z}{1+xz+z}=1\)
a) Đặt \(\frac{x}{3}=\frac{y}{5}=k\) => x = 3k ; y = 5k
Do đó x . y = 3k . 5k = 15k2 = 60
=> k2 = 4 => k = + 2
- Với k = 2 thì x = 6 ; y = 10
- Với k = - 2 thì x = -6 ; y = -10
b) Tương tự
a,Ta có: x+y= -7/6 và y+z= 1/4
=>x+y+y+z= -7/6 +1/4
=>x+z+2y= -11/12
=>1/2+2y= -11/12
=>2y= -11/12 -1/2
=>2y= -17/12
=>y= -17/24
Mà x+y=-7/6 =>x= -7/6+17/24= -11/24
x+z=1/2 =>z=1/2+11/24=23/24
Ta có: \(x+y=-\frac{7}{6};y+z=\frac{1}{4};x+z=\frac{1}{2}\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=-\frac{7}{6}+\frac{1}{4}+\frac{1}{2}\)
\(\Rightarrow2x+2y+2z=-\frac{28}{24}+\frac{6}{24}+\frac{12}{24}\)
\(\Rightarrow2\left(x+y+z\right)=-\frac{5}{12}\)
\(\Rightarrow x+y+z=-\frac{5}{12}:2\)
\(\Rightarrow x+y+z=-\frac{5}{24}\)
\(\Rightarrow\left(x+y+z\right)-\left(x+y\right)=-\frac{5}{24}+\frac{7}{6}\Rightarrow z=-\frac{5}{24}+\frac{28}{24}=\frac{23}{24}\)
\(\Rightarrow\left(x+y+z\right)-\left(y+z\right)=-\frac{5}{24}-\frac{1}{4}\Rightarrow x=-\frac{5}{24}-\frac{6}{24}=-\frac{11}{24}\)
\(\Rightarrow\left(x+y+z\right)-\left(x+z\right)=-\frac{5}{24}-\frac{1}{2}\Rightarrow y=-\frac{5}{24}-\frac{12}{24}=-\frac{17}{24}\)
Vậy \(x=\frac{23}{24};y=-\frac{17}{24};z=-\frac{11}{24}\)
Chuk pạn hok tốt!
\(\left\{{}\begin{matrix}x,y,z\ne0\\x^2.y.z=-4\\xy^2z=2\\xyz^2=-2\end{matrix}\right.\)\(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\\\left(4\right)\end{matrix}\)
(2).(3).(4) \(\left(x^2yz\right).\left(xy^2z\right)\left(xyz^2\right)=\left(x^{2+1+1}.y^{1+2+1}.z^{1+1+2}\right)=\left(xyz\right)^4=\left(-4\right).2.\left(-2\right)=8\)\(\Leftrightarrow\left[{}\begin{matrix}xyz=2\\xyz=-2\end{matrix}\right.\)\(\begin{matrix}\left(I\right)\\\left(II\right)\end{matrix}\)
TH(I)
(2) => x =-2 ;(3) => y =1;(4) => z =-1
TH(II)
(2) => x =2 ; (3) => y =-1; (4) => z =1
(x;y;z)=(-2;1;-1);(2;-1;1)