Question

I. Tìm x, biết :
1. \(\frac{x}{-15}=\frac{-60}{x}\)

2. \(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)

3. \(\frac{37-x}{x+13}=\frac{-3}{-7}\)

II. Tìm a, b, c, biết :
1. \(\frac{a}{3}=\frac{b}{4};\frac{b}{5}=\frac{c}{7}\) và \(2a+3b+c=186\)

2. \(3a=2b;7b=5c\) và \(a-b+c=32\)

3. \(\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}\) và \(2a+3b-c=50\)

Cảm ơn mấy bạn trc nghen :))))

Answer 1

I, Tìm x biết :

1.\(\frac{x}{-15}=\frac{-60}{x}\)

\(\Leftrightarrow2x=\left(-15\right).\left(-60\right)\)

\(\Leftrightarrow2x=900\)

\(\Leftrightarrow x=450\)

2. \(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)

\(\Leftrightarrow\left(x-2\right).\left(x+7\right)=\left(x-1\right).\left(x+4\right)\)

\(\Leftrightarrow x^2+7x-2x-14=x^2+4x-x-4\)

\(\Leftrightarrow5x-14=3x-4\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

Vậy : \(x=5\)

3)\(\frac{37-x}{x+13}=\frac{-3}{-7}=\frac{3}{7}\)

\(\Leftrightarrow\left(37-x\right).7=\left(x+13\right).3\)

\(\Leftrightarrow259-7x=3x+39\)

\(\Leftrightarrow220=4x\)

\(\Leftrightarrow x=55\)

Vậy : \(x=55\)

Answer 2

I.

1) \(\frac{x}{-15}=\frac{-60}{x}\)

=> \(x.x=\left(-60\right).\left(-15\right)\)

=> \(x.x=900\)

=> \(x^2=900\)

=> \(\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)

Vậy \(x\in\left\{30;-30\right\}.\)

Chúc bạn học tốt!