Ta có:

\(\frac{x^2}{6}=\frac{24}{25}\Leftrightarrow x^2.25=24.6\)

\(\Leftrightarrow x^2=\frac{24.6}{25}=\frac{144}{25}=5,76\)

\(\Leftrightarrow x=\sqrt{5,76}=2,4;x=-\sqrt{5,76}=-2,4\)

Vậy \(x=2,4\) hoặc \(x=-2,4\)

a.

\(\frac{x-1}{x-5}=\frac{6}{7}\)

\(\left(x-1\right)\times7=6\times\left(x-5\right)\)

\(7x-7=6x-30\)

\(7x-6x=-30+7\)

\(x=-23\)

b.

\(\frac{x^2}{6}=\frac{24}{25}\)

\(x^2=\frac{24}{25}\times6\)

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

\(x^2=\left(\pm\frac{12}{5}\right)^2\)

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

Vậy \(x=\frac{12}{5}\) hoặc \(x=-\frac{12}{5}\)

CÓ VẤN ĐỀ ._.

b) \(\frac{x^2}{6}=\frac{24}{25}\)

\(\Rightarrow x.25=24.6\)

\(\Rightarrow x.25=144\)

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

\(\Rightarrow x=5,76\)

Vậy x = 5,76

\(x=2,4\)

X = 3 

hok tốt

X = 2,4

mk viết nhầm

a) \(\frac{x}{6}^2=\frac{24}{25}\)

\(\Rightarrow x^2.25=6.24\)

\(\Rightarrow x^2.25=144\)

\(\Rightarrow x^2=144\div25\)

\(\Rightarrow x^2=5,76=2,4^2=\left(-2,4^2\right)\)

\(\Rightarrow x\in\left\{2,4;-2,4\right\}\)

Vậy \(x\in\left\{2,4;-2,4\right\}\)

b) \(\frac{72-x}{7}=\frac{x-40}{9}\)

\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)

\(\Rightarrow648-9x=7x-280\)

\(\Rightarrow\left(-280\right)-648\) \(=-9x-7x\)

\(\Rightarrow-928=-16x\)

\(\Rightarrow x=58\)

Vậy \(x=58\)

Ta có: \(\frac{x^2}{6}=\frac{24}{25}\)

\(\Rightarrow x^2=24.6:25=5,76\)

\(\Rightarrow x=2,4\)

Vậy x = 2,4

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

\(\Rightarrow7x-7=6x+30\)

\(\Rightarrow x=37\)

Vậy x=37

b) \(\frac{x^2}{6}=\frac{24}{25}\)

\(\Leftrightarrow x^2.25=6.24\)

\(\Leftrightarrow x^2.25=144\)

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

\(\Leftrightarrow x=\pm\sqrt{\frac{144}{25}}=\pm\frac{12}{5}\)

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

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

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow x=5\)

Đặt \(a=24-x,b=x-25\)

Khi đó pt ban đầu trở thành :

\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)

\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)

\(\Leftrightarrow30a^2+68ab+30b^2=0\)

\(\Leftrightarrow15a^2+34ab+15b^2=0\)

\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)

Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !