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LV
1
9 tháng 12 2021
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
Do đó: x=16; y=24; z=30
6 tháng 5 2023
Ta có:
`x/10=y/5 -> x/20=y/10` `(1)`
`y/2=z/3 -> y/10=z/15` `(2)`
Từ `(1)` và `(2)`
`-> x/20=y/10=z/15` `-> x/20=y/10=(4z)/60`
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
`x/20=y/10=(4z)/60=(x+4z)/(20+60)=320/80=4`
`-> x/20=y/10=z/15=4`
`-> x=20*4=80, y=10*4=40, z=15*4=60`.
N
7 tháng 5 2023
Ta có:
\(\left\{{}\begin{matrix}\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\\\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\end{matrix}\right.\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}=\dfrac{x+4z}{20+4.15}=\dfrac{320}{80}=4\)
Do đó:
\(\dfrac{x}{20}=4\Rightarrow x=80\)
\(\dfrac{y}{10}=4\Rightarrow y=40\)
\(\dfrac{z}{15}=4\Rightarrow z=60\)
AL
9 tháng 2 2019
1)
x(x-y) = \(\dfrac{3}{10}\)
=> \(x^2-xy=\dfrac{3}{10}\) (1)
y(x-y) = \(-\dfrac{3}{50}\)
=> \(xy-y^2=-\dfrac{3}{50}\) (2)
Trừ (1) cho (2), ta có :
\(x^2-xy-xy+y^2=\dfrac{3}{10}+\dfrac{3}{50}\)
\(\Rightarrow x^2-2xy+y^2=\dfrac{18}{50}=\dfrac{9}{25}\)
=> \(\left(x-y\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x-y=\dfrac{3}{5}\\x-y=-\dfrac{3}{5}\end{matrix}\right.\)
TH1
x- y = \(\dfrac{3}{5}\)
Ta có
\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{5}x=\dfrac{3}{10}\\\dfrac{3}{5}y=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{10}\end{matrix}\right.\)
TH2:
x-y=\(-\dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{3}{5}x=\dfrac{3}{10}\\-\dfrac{3}{5}y=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{5}\end{matrix}\right.\)
Vậy các cặp (x,y) thỏa mãn là (x;y) \(\in\left\{\left(\dfrac{1}{2};-\dfrac{1}{5}\right);\left(-\dfrac{1}{2};\dfrac{1}{5}\right)\right\}\)
2) \(\left(x-3\right)\left(x+\dfrac{1}{2}\right)>0\)
TH1:
\(\left\{{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{1}{2}\end{matrix}\right.\)
=> x >3
TH2:
\(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{1}{2}\end{matrix}\right.\)
=> x <\(-\dfrac{1}{2}\)
Vậy giá trị x thỏa mãn là x < -1/2 hoặc x>3
9 tháng 2 2019
1)
Từ gt,ta có : x(x - y) - y(x - y) =\(\frac{3}{10}-\frac{-3}{50}\)
(x - y)2 =\(\frac{9}{25}\)\(\Rightarrow\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}:\frac{3}{5}=\frac{1}{2}\\x=\frac{3}{10}:\frac{-3}{5}=\frac{-1}{2}\end{cases};\orbr{\begin{cases}y=\frac{-3}{50}:\frac{3}{5}=\frac{-1}{10}\\y=\frac{-3}{50}:\frac{-3}{5}=\frac{1}{10}\end{cases}}}}\)
Vậy\(x=\frac{1}{2};y=\frac{-1}{10}\) hoặc\(x=\frac{-1}{2};y=\frac{1}{10}\)
HT
2
3 tháng 9 2021
a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)
b)Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)
Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)
Ta thấy: \(3\left|x+y\right|\ge0\forall x;y\)
\(10\left|y+\dfrac{2}{3}\right|\ge0\forall y\)
\(\Rightarrow3\left|x+y\right|+10\left|y+\dfrac{2}{3}\right|\ge0\forall x;y\)
Mà: \(3\left|x+y\right|+10\left|y+\dfrac{2}{3}\right|\le0\)
nên: \(\left\{{}\begin{matrix}x+y=0\\y+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{3};y=-\dfrac{2}{3}\).