là 0

mk nhé

dien a

chu vi hình tam giác là: 5+5+5=15(cm)

A²+B²= (A+B)²-2ab

\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10\)

\(\Leftrightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=10-1-2-3-4\)

\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-166}{23}-4\right)=10-1-2-3-4\)

\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{20}+\frac{x-258}{21}=0\)

\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)

\(\Leftrightarrow x-258=0\).Do \(\frac{1}{17}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\ne0\)

\(\Leftrightarrow x=258\)

Ta có: a + b = c <=> a² + b² + 2ab = c² <=> a² + b² - c² = - 2ab

Tương tự: a² + c² - b² = - 2ac

b² + c² - a² = - 2bc

Thế vào ta được

\(\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ac}{a^2+c^2-b^2}=-\frac{ab}{2ab}-\frac{bc}{2bc}-\frac{ac}{2ac}=-6\)

=-6 ngo như bù

Ta có: \(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\)

\(\Leftrightarrow\)(x, y) = (1, 1; 1, - 1; - 1, 1; - 1, - 1)