hình vuông sao có bốn cạnh được bạn

\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)\(=\frac{3}{11}\)

1) \(\frac{17}{6}-\left(x-\frac{7}{6}\right)=\frac{7}{4}\)

\(\Rightarrow x-\frac{7}{6}=\frac{17}{6}-\frac{7}{4}\)

\(\Rightarrow x=\frac{13}{12}+\frac{7}{6}=\frac{9}{4}\)

2) \(\frac{3}{35}-\left(\frac{3}{5}-x\right)=\frac{2}{7}\)

\(\Rightarrow\)\(\frac{3}{5}-x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)

\(\Rightarrow x=\frac{3}{5}-\left(-\frac{1}{5}\right)=\frac{4}{5}\)

3) 4) Hjhj^_^^_^

\(x;y;z\ne0\). Giả thiết của đề bài:

\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{z+x}\Leftrightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{x+z}{xz}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{x}+\frac{1}{z}\Leftrightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}.\)

=> x = y = z

Do đó, M = 1.

a) x^2-9=0

=>x²-3²=0

=>(x-3)(x+3)=0 (HĐT)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

b)x^2-x=0

=>x(x-1)=0

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

c) x^2-2x=0

=>x(x-2)=0

\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

d) x^2-3x=0

=>x(x-3)=0

\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

e) 3x^2-4x=0

=>x(3x-4)=0

\(\Rightarrow\orbr{\begin{cases}x=0\\3x-4=0\Rightarrow3x=4\Rightarrow x=\frac{3}{4}\end{cases}}\)

\(A=\left[0,8.7\left(0,8\right)^2\right].\left(1,25.7-\frac{4}{5}.1,25\right)+31,64\)

  \(=0,8.\left(7+0,8\right).1,25.\left(7-0,8\right)+31,64\)

  \(=0,8.7,8.1,25.6,2+31,64\)

  \(=6,24.7,75+31,64\)

   \(=48,36+31,64=80\)

\(B=\frac{\left(1,09-0,29\right).\frac{5}{4}}{\left(18,9-16,65\right).\frac{8}{9}}=\frac{0,8.1,25}{2,25.\frac{8}{9}}=\frac{1}{2}\)

\(A:B=80:\frac{1}{2}=160\)

Vậy A gấp 160 lần B.