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\(\frac{-12}{x}=\frac{4}{13}\)
\(\Rightarrow4x=13.\left(-12\right)=-156\)
\(\Rightarrow x=-156:4=-39\)
\(\frac{-15}{72}=\frac{125}{-3y}\)
\(\Rightarrow\frac{-15}{72}=\frac{-3000}{72y}\)
\(\Rightarrow-15y=-3000\)
\(\Rightarrow y=-3000:-15=200\)
\(\frac{49}{21}=\frac{-14z}{-7}\)
\(\Rightarrow\frac{49}{21}=\frac{42z}{21}\)
\(\Rightarrow49=42z\)
\(\Rightarrow z=\frac{49}{42}\)
a) \(\frac{3x-2}{1\frac{2}{5}}=\frac{2\frac{3}{7}}{2\frac{3}{5}}\Rightarrow3x-2=1.\frac{2}{5}.\frac{2\frac{3}{7}}{2\frac{3}{5}}=\frac{7}{5}.\frac{17}{7}.\frac{5}{13}\)
\(3x-2=\frac{17}{13}\Rightarrow3x=2+\frac{17}{13}=\frac{43}{13}\Rightarrow x=\frac{43}{39}.\)
b) \(\frac{x}{0,16}=\frac{9}{x}\Rightarrow x^2=9.0,16=1,44\Rightarrow x=\pm1,2\)
Bài 2:
a: =>x/4=1/8
hay x=1/2
b: \(\Leftrightarrow\dfrac{x}{-5}=\dfrac{11}{6}\)
hay x=-55/6
c: \(\Leftrightarrow\dfrac{-3.5}{x}=\dfrac{4.25}{8}\)
hay x=-112/17
a, \(\frac{x-6}{x+4}=\frac{2}{7}\Rightarrow7x-42=2x+8\)ĐK : \(x\ne-4\)
\(\Leftrightarrow5x=50\Leftrightarrow x=10\)(tm)
b, \(\left(x+5\right):2\frac{1}{2}=\frac{40}{x+5}\)ĐK : \(x\ne-5\)
\(\Leftrightarrow\frac{5\left(x+5\right)}{2}=\frac{40}{x+5}\Rightarrow5\left(x+5\right)^2=80\Leftrightarrow\left(x+5\right)^2=16\)
TH1 : \(x+5=4\Leftrightarrow x=-1\)
TH2 : \(x+5=-4\Leftrightarrow x=-9\)
a)\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
b)\(\frac{37-x}{x+13}=\frac{3}{7}\)\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow10x=220\)
\(\Rightarrow x=22\)
\(\text{a)}\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Leftrightarrow3\left(x+13\right)=7\left(37-x\right)\)
\(\Leftrightarrow3x+39=259-7x\)
\(\Leftrightarrow3x+7x=259-39\)
\(\Leftrightarrow10x=220\)
\(\Leftrightarrow x=22\)
\(\text{b)}\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Leftrightarrow\left(x+4\right)^2=100\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)
a) - 0,52 : x = - 9,36 : 16,38
- 0,52 : x = \(-\frac{4}{7}\)
\(x=\left(-0,52\right):\left(-\frac{4}{7}\right)=\frac{91}{100}=0,91\)
b) \(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)
\(\frac{x}{1,61}=\frac{34}{23}\)
\(x=\frac{34}{23}.1,61=2,38\)
b) Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+3y+1+5y+1+7y}{12+5x+4x}=\frac{3+15y}{12+5x+4x}=\frac{3\left(1+5y\right)}{2.3.2+5x+4x}=\frac{1+5y}{4+9x}=\frac{1+5y}{5x}\)<=> 4 + 9x = 5x
....
a/ Từ giả thiêt ta có \(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\Leftrightarrow\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\). Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\)
\(\Rightarrow\begin{cases}x=15k\\y=20k\\z=40k\end{cases}\)
Theo đề bài : \(xy=1200\Leftrightarrow15k.20k=1200\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Tới đây dễ rồi nhé :)
b/ \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\Leftrightarrow\frac{1+5y}{5}=\frac{1+7y}{4}\Leftrightarrow\frac{7+35y}{35}=\frac{5+35y}{20}=\frac{7+35y-5-35y}{35-20}=\frac{2}{15}\)
\(\Rightarrow y=-\frac{1}{15}\)
Thay y vào \(\frac{1+3y}{12}=\frac{1+5y}{5x}\) tìm được x = 2
=> (72 - x) . 9 = (x - 40) . 7
=> 648 - 9x = 7x - 280
=> (-280) - 648 = -9x - 7x
-928 = -16x
=> x = 58