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a,\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Leftrightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)=3
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:
\(M=\frac{2020x^3+2020y^3+2020z^3}{x^3+y^3+z^3}+\frac{2021x^5+2021y^5}{x^5+y^5}\)
\(\Rightarrow M=\frac{2020\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}+\frac{2021\left(x^5+y^5\right)}{x^5+y^5}\)
\(\Rightarrow M=2020+2021=4041\)
a) Cộng cả 3 đẳng thức trên ta có:
2(x + y + z) = 1/2 +1/3 + 1/4 = 13/12 => x + y + z = 13/24 (*)
z = 13/24 - 1/2 = 1/24
x = 13/24 - 1/3 = 5/24
y = 13/24 - 1/4 = 7/24.
b) Nhân cả 3 đẳng thức ta có: x2y2z2 = 1/16 => xyz = 1/4 hoặc -1/4
- Nếu xyz = 1/4 thì: z = -1/2; x = 1/2; y = -1
- Nếu xyz = -1/4 thì: z = 1/2; x = -1/2; y = 1
\(\left\{{}\begin{matrix}xy=\dfrac{3}{5}\\yz=\dfrac{4}{5}\\zx=\dfrac{3}{4}\end{matrix}\right.\Rightarrow x^2y^2z^2=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}=\dfrac{9}{25}\)
\(\Rightarrow xyz=\pm\dfrac{3}{5}\)
+) \(xyz=\dfrac{3}{5}\Rightarrow\left\{{}\begin{matrix}z=1\\x=\dfrac{3}{4}\\y=\dfrac{4}{5}\end{matrix}\right.\)
+) \(xyz=\dfrac{-3}{5}\Rightarrow\left\{{}\begin{matrix}z=-1\\x=\dfrac{-3}{4}\\y=\dfrac{-4}{5}\end{matrix}\right.\)
Vậy...
\(\text{Ta có : }xy=\dfrac{3}{5}\\ yz=\dfrac{4}{5}\\ zx=\dfrac{4}{4}\\ \Rightarrow xy\cdot yz\cdot zx=\dfrac{3}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}\\ \Rightarrow x^2\cdot y^2\cdot z^2=\dfrac{9}{25}\Rightarrow\left(xyz\right)^2=\dfrac{9}{25}\\ \Rightarrow xyz=\dfrac{-3}{5}\text{hoặc : }\\ xyz=\dfrac{3}{5}\)
\(\text{+) Xét }xyz=-\dfrac{3}{5}\Leftrightarrow\left\{{}\begin{matrix}x\cdot\left(yz\right)=-\dfrac{3}{5}\\y\cdot\left(xz\right)=-\dfrac{3}{5}\\z\cdot\left(xy\right)=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\cdot\dfrac{4}{5}=-\dfrac{3}{5}\\y\cdot\dfrac{3}{4}=-\dfrac{3}{5}\\z\cdot\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=-\dfrac{4}{5}\\z=-1\end{matrix}\right.\)
\(\text{+) Xét }xyz=\dfrac{3}{5}\Leftrightarrow\left\{{}\begin{matrix}x\cdot\left(yz\right)=\dfrac{3}{5}\\y\cdot\left(xz\right)=\dfrac{3}{5}\\z\cdot\left(xy\right)=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\cdot\dfrac{4}{5}=\dfrac{3}{5}\\y\cdot\dfrac{3}{4}=\dfrac{3}{5}\\z\cdot\dfrac{3}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{4}{5}\\z=1\end{matrix}\right.\)
Vậy \(x;y;z=-\dfrac{3}{4};-\dfrac{4}{5};-1\) hoặc \(x;y;z=\dfrac{3}{4};\dfrac{3}{5};1\)
Theo bài ra: x.y=\(\frac{3}{5}\)(1)
y.z=\(\frac{4}{5}\)(2)
z.x=\(\frac{3}{4}\)(3)
Ta có: x.y.y.z.z.x=\(\frac{3}{5}.\frac{4}{5}.\frac{3}{4}\)\(\Leftrightarrow\)(x.y.z)\(^2\)=\(\frac{9}{25}\)\(\Rightarrow\)x.y.z=\(\frac{3}{5}\)
Từ (1), ta có:x.y=\(\frac{3}{5}\), mà x.y.z=\(\frac{3}{5}\)\(\Rightarrow\)z=1
Từ (2), ta có:y.z=\(\frac{4}{5}\), mà x.y.z=\(\frac{3}{5}\)\(\Rightarrow\)x=\(\frac{3}{4}\)
Ta có: x.y.z=\(\frac{3}{5}\), mà z=1;x=\(\frac{3}{4}\)\(\Rightarrow\)y=\(\frac{4}{5}\)
Ta có x.y.y.z.z.x = 3/5.4/5.3/4
(=) (x.yz)^2 = 9/25
mà (x.yz)^2 = (3/5)^2
=> x.y.z =3/5
Tới đây bạn chia cho các đẳng thức đã cho và tìm được ra x;y;z
Vậy z=1
x=3/4
y=4/5
\(\left(xy\right):\left(yz\right)=\frac{2}{3}:0,6\Rightarrow\frac{x}{z}=\frac{10}{9}\)=> \(x=\frac{10}{9}z\Rightarrow\frac{10}{9}z.z=0,625\Rightarrow z^2=\frac{9}{16}\Rightarrow z=\pm\frac{3}{4}\)
\(\left(yz\right):\left(zx\right)=0,6:0,625\Rightarrow\frac{y}{x}=\frac{24}{25}\)
Với z=3/4 => x, y
Với z=-3/4 => x,y
Câu b làm tương tự nhé :)
