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Bài 1: \(x\).(\(x-y\)) = \(\dfrac{3}{10}\) và y(\(x-y\)) = - \(\dfrac{3}{50}\)
\(x\)(\(x\) - y) - y(\(x\) - y) = \(\dfrac{3}{10}\) - ( - \(\dfrac{3}{50}\))
(\(x-y\)).(\(x-y\)) = \(\dfrac{3}{10}\) + \(\dfrac{3}{50}\)
(\(x-y\))2 = \(\dfrac{15}{50}\) + \(\dfrac{3}{50}\)
(\(x\) - y)2 = \(\dfrac{9}{25}\) = (\(\dfrac{3}{5}\))2
\(\left[{}\begin{matrix}x-y=-\dfrac{3}{5}\\x-y=\dfrac{3}{5}\end{matrix}\right.\)
TH1 \(x-y=-\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\left(-\dfrac{3}{5}\right)=\dfrac{3}{10}\\y.\left(-\dfrac{3}{5}\right)=-\dfrac{3}{50}\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\left(-\dfrac{3}{5}\right)=\dfrac{-1}{2}\\y=-\dfrac{3}{50}:\left(-\dfrac{3}{5}\right)=\dfrac{1}{10}\end{matrix}\right.\)
TH2: \(x-y=\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\dfrac{3}{5}=\dfrac{3}{10}\\y.\dfrac{3}{5}=-\dfrac{3}{50}\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\dfrac{3}{5}=\dfrac{1}{2}\\y=-\dfrac{3}{50}:\dfrac{3}{5}=-\dfrac{1}{10}\end{matrix}\right.\)
Vậy (\(x;y\) ) = (- \(\dfrac{1}{2}\); \(\dfrac{1}{10}\)); (\(\dfrac{1}{2}\); - \(\dfrac{1}{10}\))
Áp dụng tính chất dãy tỉ số bằng nhau ta có : a-b/x = b-c/y = a-c/z = a-b+b-c+c-a/x+y+z = 0
=> a-b=0 ; b-c=0 ; c-a=0
=> a=b=c
Tk mk nha
Vì \(\left|2x+1\right|\ge0;\left|x+y-\frac{1}{2}\right|\ge0\)
Mà \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\Rightarrow\orbr{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{4}\end{cases}}\)(1)
Thế (1) vào A
\(\Rightarrow A=4.\left(-\frac{1}{2}\right)^3.\left(\frac{1}{4}\right)^2-\frac{1}{4}.\left(-\frac{1}{2}\right)+2.\frac{1}{4}-5\)
\(\Rightarrow A=-\frac{1}{2}+\frac{1}{8}+\frac{1}{2}-5\)
\(\Leftrightarrow A=\frac{1}{8}-5=\frac{1}{8}-\frac{40}{8}=-\frac{39}{8}\)
\(\text{Ta có: }\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{x.\left(x+1\right)}=\frac{13}{90}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{13}{90}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
=> x = 17
a, \(\frac{1}{x}=\frac{1}{6}+\frac{y}{3}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{6}+\frac{2y}{6}=\frac{1+2y}{6}\)
\(\Rightarrow1\cdot6=x\cdot\left(1+2y\right)\)
\(\Rightarrow x\left(1+2y\right)=6\)
\(\Rightarrow x;1+2y\inƯ\left(6\right)=\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
ta có bảng :
| x | -1 | 1 | -2 | 2 | -3 | 3 | -6 | 6 |
| 1+2y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 1 |
| y | loại | loại | 2 | -1 | loại | loại | 1 | 0 |
vậy_
phần b tương tự
\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
Vậy \(P=\frac{3}{11}\)
Bài 1:
\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{1}{7}+\frac{11}{13}}\)
\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}-\frac{11}{3}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{13}\right)}{11.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
Bài 2:
a) \(\left(x+1\right)\left(x-2\right)< 0\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)=0\left(\text{loại}\right)\\\left(x-2\right)=0\end{cases}}\Rightarrow x=2\)
nhanh hộ mik vs.mai miik phải nộp bài r.giúp mik đi