Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{a-x}{b-y}=\dfrac{a}{b}\)
\(\Rightarrow\dfrac{a-x}{a}=\dfrac{b-y}{b}\)
\(\Rightarrow1-\dfrac{x}{a}=1-\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{a}{b}\)
Bài 4:
a) \(\dfrac{2.7.13}{26.35}=\dfrac{2.7.13}{13.2.7.5}=\dfrac{1}{5}\)
b) \(\dfrac{23.5-23}{4-27}=\dfrac{23.\left(5-1\right)}{-23}=\dfrac{23.4}{-23}=-4\)
c) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)
Bài này dễ nên nãy không có định làm, nhưng mà nghĩ lại thôi làm vậy:Đ
a/ \(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{2}{4}\)
Ta có: \(\dfrac{-4}{y}=\dfrac{2}{4}\Rightarrow y=\dfrac{-4.4}{2}=-8\)
\(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{-4}{-8}=\dfrac{1}{2}\Rightarrow x=\dfrac{-2.1}{2}=-1\)
b/\(\dfrac{2}{x}=\dfrac{y}{-3}\Rightarrow xy=-3.2=-6\)
\(\Rightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Ta có bảng giá trị của x, y như sau:
| x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
| y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 1 |
c/ \(\dfrac{x+1}{2}=\dfrac{8}{x+1}\Rightarrow\left(x+1\right)^2=2.8=16\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=4^2\\\left(x+1\right)^2=\left(-4\right)^2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
| x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
| y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 1 | 1 | -1 | 3 | -3 |
| y + 1 | 3 | -3 | 1 | -1 |
| x | 2 | 0 | 4 | -2 |
| y | 2 | -4 | 0 | -2 |
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
\(\dfrac{a-x}{b-y}=\dfrac{a}{b}\)
\(\Rightarrow\left(a-x\right).b=\left(b-y\right).a\)
\(\Rightarrow ab-xb=ba-ya\)
\(\Rightarrow xb=ya\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{x}{y}\) (đpcm)
a) Ta có: \(\dfrac{3+x}{7+y}=\dfrac{3}{7}\)
\(\Leftrightarrow\dfrac{x+3}{3}=\dfrac{y+7}{7}\)
mà x+y=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x+3}{3}=\dfrac{y+7}{7}=\dfrac{x+y+3+7}{3+7}=\dfrac{20+10}{10}=3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x+3}{10}=3\\\dfrac{y+7}{7}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3=30\\y+7=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=14\end{matrix}\right.\)
Vậy: x=27; y=14
a) 2 cái đều nhân lại lớn hơn 0, x.y = xy
=> đfcm
b) 2 cái giống nhau nên=> đfcm
Đúng thì like giúp mik nha bạn. Thx bạn
Ta có: \(\frac{a-x}{b-y}=\frac{a}{b}\Rightarrow\left(a-x\right)b=\left(b-y\right)a\)
\(\Rightarrow ab-bx=ab-ay\Rightarrow bx=ay\)
\(\Rightarrow\frac{x}{y}=\frac{a}{b}\left(ĐPCM\right)\)