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a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
=>\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{-3x-4y+5z+3-12-25}{-6-16+30}=2\)
=>x-1=4; y+3=8; z-5=12
=>x=5; y=5; z=17
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
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) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
a) Ta có :
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{2y}{6}=\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{5-2y}{6}=\dfrac{4}{x}\)
\(\Leftrightarrow\left(5-2y\right)x=6.4=24\)
Vì \(x,y\in N\Leftrightarrow5-2y\in N;5-2y;x\inƯ\left(24\right)\)
Ta có bảng :
Vậy ...
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{2y}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)
\(\Rightarrow x\left(5-2y\right)=24\)
\(\Rightarrow x;5-2y\inƯ\left(24\right)\)
Xét ước là xong
\(3x-xy-4y+12=17\)
\(\Rightarrow x\left(3-y\right)+4\left(3-y\right)=17\)
\(\Rightarrow\left(x+4\right)\left(3-y\right)=17\)
\(\Rightarrow x+4;3-y\inƯ\left(17\right)\)
\(Ư\left(17\right)=\left\{\pm1;\pm17\right\}\)
Xét ước