Timf x,y,z
x.y=-8/5 , y.z=3/4 , z.x=-3/10
tìm x,y,z biết x/8=y/3=z/10 và (x.y)+(y.z)+(z.x)=206
Đặt \(k=\frac{x}{8}=\frac{y}{3}=\frac{z}{10}\)
Ta có: \(x=8k;y=3k;z=10k\) (*)
Thay vào đẳng thức \(xy+yz+zx=206\) ta được:
\(8k.3k+3k.10k+10k.8k=206\)
\(\Leftrightarrow24k^2+30k^2+80k^2=206\)
\(\Leftrightarrow24k^2+30k^2+80k^2=206\)
\(\Rightarrow k=\pm\sqrt{\frac{103}{67}}\)
Thay k vào (*) tính được x, y, z
tìm các số hữu tỉ x ,y,z thảo mãn
x.y=-2/5 ; y.z=3/4 ; z.x=-3/10
\(\Rightarrow\left(x.y.z\right)^2=\frac{-2}{5}.\frac{3}{4}.\frac{-3}{10}\)
\(\Rightarrow\left(x.y.z\right)^2=\frac{18}{200}=\frac{9}{100}\)
\(\Rightarrow x.y.z=\frac{3}{10}\)
\(\Rightarrow z=\frac{3}{-4}\)
\(\Rightarrow x=\frac{2}{5}\)
\(\Rightarrow y=-1\)
Tìm các số hữu tỉ x,y,z
x (x+y+z) = -12 ; y (y+x+z) = 18 ; z (z+y+x) = 30
\(\frac{x}{3}=\frac{y}{5};\frac{y}{6}=\frac{z}{7}\)và 3x + y - 2z = 42
x.y = z; y.z = 4x ; z.x = 9y
x.y = \(\frac{3}{5};y.z=\frac{4}{5};z.x=\frac{3}{4}\)
Tìm x
a) x+y=1/3 b) x.y=3/5
y+z=-1/4 y.z=4/5
z+x=1/5 z.x=3/4
a)Ta có:
\(\left\{{}\begin{matrix}x+y=\frac{1}{3}\\y+z=\frac{-1}{4}\\z+x=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{3}+\frac{-1}{4}+\frac{1}{5}\)
\(\Rightarrow2\left(x+y+z\right)=\frac{17}{60}\)
\(\Rightarrow x+y+z=\frac{17}{60}:2=\frac{17}{120}\)
\(\Rightarrow\left\{{}\begin{matrix}z=\frac{-23}{120}\\x=\frac{47}{120}\\y=\frac{-7}{120}\end{matrix}\right.\)
b)Ta có:
\(\left\{{}\begin{matrix}xy=\frac{3}{5}\\yz=\frac{4}{5}\\zx=\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow xyyzzx=\frac{3}{5}.\frac{4}{5}.\frac{3}{4}=\frac{9}{25}\)
\(\Rightarrow\left(xyz\right)^2=\frac{9}{25}\Rightarrow\left[{}\begin{matrix}xyz=\frac{3}{5}\\xyz=-\frac{3}{5}\end{matrix}\right.\)
TH1: \(xyz=\frac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=1\\x=\frac{3}{4}\\y=\frac{4}{5}\end{matrix}\right.\)
TH2:
\(xyz=-\frac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=-1\\x=-\frac{3}{4}\\y=-\frac{4}{5}\end{matrix}\right.\)
Tìm x ; y ; z :
\(x.y=\frac{3}{5}\) ; \(y.z=\frac{4}{5}\) ; \(z.x=\frac{3}{4}\)
\(\left\{{}\begin{matrix}xy=\dfrac{3}{5}\\yz=\dfrac{4}{5}\\zx=\dfrac{3}{4}\end{matrix}\right.\Rightarrow x^2y^2z^2=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}=\dfrac{9}{25}\)
\(\Rightarrow xyz=\pm\dfrac{3}{5}\)
+) \(xyz=\dfrac{3}{5}\Rightarrow\left\{{}\begin{matrix}z=1\\x=\dfrac{3}{4}\\y=\dfrac{4}{5}\end{matrix}\right.\)
+) \(xyz=\dfrac{-3}{5}\Rightarrow\left\{{}\begin{matrix}z=-1\\x=\dfrac{-3}{4}\\y=\dfrac{-4}{5}\end{matrix}\right.\)
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 \(x;y;z=\dfrac{3}{4};\dfrac{3}{5};1\)
\(\left\{{}\begin{matrix}x.y=\dfrac{3}{5}\\y.z=\dfrac{4}{5}\\z.x=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow x.y.y.z.z.x=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}\)
\(\Rightarrow\left(x.y.z\right)^2=\dfrac{9}{25}\)
\(\Rightarrow xyz=\pm\dfrac{9}{25}\)
Làm tiếp
Tìm x ; y ; z :
\(x.y=\frac{3}{5}\) ; \(y.z=\frac{4}{5}\) ; \(z.x=\frac{3}{4}\)
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=\(\frac{4}{5}\)
Tìm x ; y ; z :
\(x.y=\frac{3}{5}\) ; \(y.z=\frac{4}{5}\) ; \(z.x=\frac{3}{4}\)
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
Tìm x,y,z:
x.y=1/2 ; y.z=1/5 ; z.x=27/10
tìm x,y,x biết
a)\(\frac{x}{3}=\frac{y}{4};\frac{y}{3}=\frac{z}{5}\)và 2x-3y+z=6
b)\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)và x.y+y.z+z.x=64
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