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1) Ta có:
\(\frac{1+2y}{18}=\frac{1+4y}{24}\)\(\Rightarrow\left(1+2y\right).24=\left(1+4y\right).18\)
=> 24 + 48y = 18 + 72y
=> 72y - 48y = 24 - 18
=> 24y = 6
\(\Rightarrow y=\frac{6}{24}=\frac{1}{4}\)
Thay \(y=\frac{1}{4}\) vào đề bài ta có:
\(\frac{1+2.\frac{1}{4}}{18}=\frac{1+6.\frac{1}{4}}{6x}\)
\(\Rightarrow\frac{1+\frac{1}{2}}{18}=\frac{1+\frac{3}{2}}{6x}\)
\(\Rightarrow\frac{3}{2}.\frac{1}{18}=\frac{5}{2}:6x\)
\(\Rightarrow\frac{1}{12}=\frac{5}{2}:6x\)
\(\Rightarrow6x=\frac{5}{2}:\frac{1}{12}=\frac{5}{2}.12=30\)
=> x = 30 : 6 = 5
Vậy \(x=5;y=\frac{1}{4}\)
2) Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(x+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2.\left(x+y+z\right)}{x+y+z}=2\)
\(=\frac{1}{x+y+z}\) (theo đề bài)
\(\Rightarrow x+y+z=\frac{1}{2}\)
Ta có: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=2\)
\(\Rightarrow\frac{y+z+1}{x}+1=\frac{x+z+2}{y}+1=\frac{x+y-3}{z}+1=2+1\)
\(\Rightarrow\frac{x+y+z+1}{x}=\frac{x+y+z+2}{y}=\frac{x+y+z-3}{z}=3\)
\(\Rightarrow\frac{\frac{1}{2}+1}{x}=\frac{\frac{1}{2}+2}{y}=\frac{\frac{1}{2}-3}{z}=3\)
\(\Rightarrow\frac{3}{2}:x=\frac{5}{2}:y=\frac{-5}{2}:z=3\)
\(\Rightarrow\begin{cases}x=\frac{3}{2}:3=\frac{1}{2}\\y=\frac{5}{2}:3=\frac{5}{6}\\z=\frac{-5}{2}:3=\frac{-5}{6}\end{cases}\)
Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=\frac{-5}{6}\)
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
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\(x=\frac{-1}{2};y=\frac{1}{10}\)
Tìm x, y, z biết:
a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và 2x + 3y + z = 17
Giải
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{z}{4}\) và 2x + 3y + z = 17
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{z}{4}=\dfrac{2x+3y+z}{4+9+4}=\dfrac{17}{17}=1\)
\(\dfrac{x}{2}=1\Rightarrow x=2\)
\(\dfrac{y}{3}=1\Rightarrow y=3\)
\(\dfrac{z}{4}=1\Rightarrow z=4\)
Vậy...
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và (x - y)2 + (y - z)2 = 2
Giải
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2}{\left(2-3\right)^2+\left(3-4\right)^2}=\dfrac{2}{2}=1\)
\(\dfrac{x}{2}=1\Rightarrow x=2\)
\(\dfrac{y}{3}=1\Rightarrow y=3\)
\(\dfrac{z}{4}=1\Rightarrow z=4\)
Vậy...
chịu thôi
x3 + x2y - y + x + x3y2 - x3 + x2y
= x3 - x3 + x2y + x2y - y + x + x3y2
= 2x2y - y + x + x3y2