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1) a) Ta có: \(\frac{x}{-15}=\frac{-60}{x}\) \(\Rightarrow x^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow x=30\)
b) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\) \(\Rightarrow x.\left(-x\right)=\left(-2\right).\frac{8}{25}\)
\(\Rightarrow x.\left(-x\right)=\frac{-16}{25}\)
\(\Rightarrow x.\left(-x\right)=\left(\frac{-4}{5}\right).\frac{4}{5}\)
Vậy \(x=\frac{4}{5}\)
2) a) \(3,8: \left(2x\right)=\frac{1}{4}:2\frac{2}{3}\)
\(\Rightarrow3,8: \left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=\frac{3}{32}:3,8=\frac{15}{608}\)
\(x=\frac{15}{608}:2=\frac{15}{1216}\)
Vậy \(x=\frac{15}{1216}\)
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(\Rightarrow\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}.3=20\)
\(\Rightarrow x=20:0,25=80\)
Vậy x = 80
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(\Rightarrow\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Leftrightarrow0,75x=\frac{1}{250}.0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
Vậy \(x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{5}{3}.\frac{2}{3}=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}:0,1=\frac{100}{9}\)
Vậy \(x=\frac{100}{9}\)
a) \(\frac{x}{-15}=\frac{-60}{x}\Leftrightarrow x.x=-15.\left(-60\right)\Leftrightarrow x^2=900\Leftrightarrow x^2=\orbr{\begin{cases}30^2\\\left(-30\right)^2\end{cases}}\Leftrightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
a, \(\frac{x-a}{m}=\frac{y-b}{n}=\frac{x+y-a-b}{m+n}\)
\(\Rightarrow\hept{\begin{cases}\frac{x-a}{m}=\frac{x+y-a-b}{m+n}\\\frac{y-b}{n}=\frac{x+y-a-b}{m+n}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{m\left(x+y-a-b\right)}{m+n}+a\\y=\frac{n\left(x+y-a-b\right)}{m+n}+b\end{cases}}}\)
b, Áp dụng TCDTSBN với 2 tỉ số đầu ta có:
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{x}{8}=\frac{xy}{200}\)
\(\Rightarrow8xy-200x=0\Rightarrow8x\left(y-25\right)=0\Rightarrow\orbr{\begin{cases}x=0\\y=25\end{cases}}\)
Với x=0 thì \(\frac{0-y}{3}=\frac{0+y}{13}=0\Rightarrow y=0\)(t/m)
Với y=25 thif \(\frac{x-25}{3}=\frac{x+25}{13}\Rightarrow13\left(x-25\right)=3\left(x+25\right)\Rightarrow13x-325=3x+75\Rightarrow x=40\)(t/m)
Vậy các cặp (x;y) là (0;0);(40;25)
Nguyễn Hải Đăng chắc bn giỏi nói ng ta ngu :((
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y-x-y}{3-13}=\frac{-2y}{-10}=\frac{y}{5}\)
\(\Rightarrow\frac{y}{5}=\frac{xy}{200}\Rightarrow200y=5xy\Rightarrow\frac{200y}{5y}=x\Rightarrow x=40\)
\(\frac{x-y}{3}=\frac{y}{5}=\frac{40-y}{3}=\frac{y}{5}\Rightarrow5.\left(40-y\right)=3y\Rightarrow200-5y=3y\)
\(\Rightarrow200=8y\Rightarrow y=25\)
Vậy x=40, y=25
Ta có :
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{25x}{200}=\frac{xy}{200}\)
\(\Rightarrow25x=xy\Rightarrow y=25\)
\(\Rightarrow\frac{x-25}{3}=\frac{x+25}{13}\)
\(\Leftrightarrow13x-325=3x+75\)
\(\Leftrightarrow10x=400\Rightarrow x=40\)
Vậy \(x=40;y=25\)
Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{xy}{200}\left(1\right)\)
\(\Rightarrow\frac{x-y}{3}=\frac{x+y}{13}=\frac{xy}{200}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{x}{8}\left(2\right)\)
Từ (1) và (2) => \(\frac{x}{8}=\frac{xy}{200}\Rightarrow8xy=200x\)
\(\Leftrightarrow8xy-200x=0\)
\(\Leftrightarrow8x.\left(y-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}8x=0\\y-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\y=25\end{cases}}}\)
* Nếu x = 0 thì \(\frac{0-y}{3}=\frac{0+y}{13}=0\Rightarrow y=0\)
* Nếu y = 25 thì \(\frac{x-25}{3}=\frac{x+25}{13}\)
\(\Leftrightarrow13.\left(x-25\right)=3.\left(x+25\right)\)
\(\Leftrightarrow13x-325=3x+75\)
\(\Rightarrow13x-3x=75+325=400\)
\(\Rightarrow10x=400\)
\(\Rightarrow x=40\)
Vậy x =0 thì y =0
x =40 thì y = 25
Theo t/c dãy tỉ số=nhau:
(x-y)/3=(x+y)/13=(x-y+x+y)/(3+13)=2x/16=x/8
Khi đó x/8=xy/200=>200x=8xy=>200=8y=>y=25
=>x=40( bn thay y vào đề bài là tính đc x)
Vậy (x;y)=(40;25)
Theo TCDTSBN:
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{x}{8}\)
=>\(\frac{x}{8}=\frac{xy}{200}\)
=>\(\frac{x}{xy}=\frac{8}{200}\)=>\(\frac{1}{y}=\frac{8}{200}\)=>\(y=\frac{200}{8}=25\)
Khi đó ta có:\(\frac{x-25}{3}=\frac{x+25}{13}\)
=>13(x-25)=3(x+25)
=>13x-325=3x+75
=>13x-3x=75+325=>10x=400=>x=40
Vậy (x;y)=(40;25)
x=y=0 sẽ thỏa mãn biểu thức trên