Question

            Tìm x trong các tỉ lệ thức sau

a,\(\frac{x-1}{x+5}=\frac{6}{7}\)( \(x\ne-5\))

b, \(\frac{^{x^2}}{6}=\frac{24}{25}\)

c, \(\frac{x-2}{x-4}=\frac{x+4}{x+7}\)

Answer 1

a) \(\frac{x-1}{x+5}=\frac{6}{7}\)

7( x - 1 ) = 6( x + 5 )

7x - 7 = 6x - 30

7x - 6x = -30 + 7

x = -23

b) \(\frac{x^2}{6}=\frac{24}{25}\)

\(x^2.25=6.24\)

\(x^2.25=144\)

\(x^2=\frac{144}{25}\)

\(x=\sqrt{\frac{144}{25}}\)

\(x=\frac{12}{5}\)

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

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

\(x^2+7x-2x-14=x^2-4x+4x-16\)

\(x^2+5x-14=x^2^{ }-16\)

\(x^2-x^2+5x=-16+14\)

\(5x=-2\)\(x=\frac{-2}{5}\)