X=-3/4

Đáng lẽ (a-b)²/ (a-d)² là \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)chứ ? Có chép sai đề không vậy ?

Ta có: 2x = 3y = 5z => \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

        \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=150\)

=> \(\hept{\begin{cases}\frac{x}{\frac{1}{2}}=150\\\frac{y}{\frac{1}{3}}=150\\\frac{z}{\frac{1}{5}}=150\end{cases}}\) => \(\hept{\begin{cases}x=150\cdot\frac{1}{2}=75\\y=150\cdot\frac{1}{3}=50\\z=150\cdot\frac{1}{5}=30\end{cases}}\)

Vậy ...

Sửa chút, chỗ mẫu 11c + 3b thành 11c +3d
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{c}=\frac{b}{d}=\frac{11a}{11c}=\frac{3b}{3d}=\frac{11a+3b}{11c+3d}\\\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{11b}{11d}=\frac{3a-11b}{3c-11d}\end{cases}}\)
\(\Rightarrow\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)
Vậy \(\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\)

\(\frac{x}{4}=\frac{y}{5}\Rightarrow\left(\frac{x}{4}\right)^2=\left(\frac{y}{5}\right)^2\Rightarrow\frac{x^2}{16}=\frac{y^2}{25}=\frac{x^2}{16}=\frac{3y^2}{3.25}=\frac{x^2}{16}=\frac{3y^2}{75}\)
Áp dụng tính chất DTSBN
\(\frac{x^2}{16}=\frac{y^2}{25}=\frac{x^2}{16}=\frac{3y^2}{75}=\frac{x^2-3y^2}{16-75}=\frac{-59}{-59}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{x^2}{16}=1\Rightarrow x^2=1.16=16=4^2\\\frac{y^2}{25}=1\Rightarrow y^2=1.25=25=5^2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x\in\left\{4;-4\right\}\\y\in\left\{5;-5\right\}\end{cases}}\)
Vậy ...

\(\frac{3}{7}x-\frac{1}{2}=\frac{1}{7}\)

\(\frac{3}{7}x=\frac{1}{7}+\frac{1}{2}\)

\(\frac{3}{7}x=\frac{9}{14}\)

\(x=\frac{3}{2}\)

\(\frac{3}{7}.x-\frac{1}{2}=\frac{1}{7}\)

\(\frac{3}{7}.x=\frac{1}{7}+\frac{1}{2}\)

\(\frac{3}{7}.x=\frac{9}{14}\)

        \(x=\frac{9}{14}:\frac{3}{7}\)

        \(x=\frac{3}{2}\)

 vậy \(x=\frac{3}{2}\) 

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