Lời giải:

Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$

$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$

$\Rightarrow x=y=z$.

Do đó:

$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$

Lời giải:

Từ điều kiện đề bài suy ra:
$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$

$\Rightarrow (\frac{x}{y})^3=(\frac{y}{z})^3=(\frac{z}{x})^3=\frac{x}{y}.\frac{y}{z}.\frac{z}{x}=1$
$\Rightarrow \frac{x}{y}=\frac{y}{z}=\frac{z}{x}=1$

$\Rightarrow x=y=z$.

Do đó:

$\frac{(x+y+z)^{2022}}{x^{337}.y^{674}.z^{1011}}=\frac{(3x)^{2022}}{x^{337}.x^{674}.x^{1011}}=\frac{3^{2022}.x^{2022}}{x^{2022}}=3^{2022}$

\(\frac{1}{337.291}+\frac{583}{291}-\frac{2}{337.291}=\frac{583}{291}-\frac{1}{337.291}=\frac{583.337-1}{337.291}=\frac{169652}{98067}\)

\(N=\frac{1}{337}+\frac{583}{291}-\frac{290}{291\cdot337}-\frac{2}{291\cdot337}=\frac{291-196471}{337.291}-\frac{292}{337\cdot291}=\frac{-196472}{337.291}\)

Tính chậm đc ko :v

=-137496 :)

=2022.19731011. (-32-2. 2021)

=?

=?

=? :v

\(A=335^{336}+336^{337}+337^{338}\)
\(=\left(...5\right)+\left(...6\right)+\left(...1\right)\)
\(=...2\)
Vậy A chia 5 dư 2

=337+829+145-829-337=145

=337+829+145-829-337

=145

=337+829+145-829-337=145