4 / 2 2 /atan2(-im(m), -re(m))\ 4 / 2 2 /atan2(-im(m), -re(m))\
\/ 3 *\/ im (m) + re (m) *cos|| i*\/ 3 *\/ im (m) + re (m) *sin||
\ 2 / \ 2 /
n1 = - -
3 3
n2 = +
/ / \\ / / \\
/ 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)|| / 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)||
/ / \ / \ |atan2| - + || / / \ / \ |atan2| - + ||
/ |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /| / |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /|
n3 = - 4 / | - | + | + | *cos|| - i*4 / | - | + | + | *sin||
\/ \ 6 6 / \ 6 6 / \ 2 / \/ \ 6 6 / \ 6 6 / \ 2 /
n4 = 4 / | - | + | + | *cos|| + i*4 / | - | + | + | *sin||
/ / \ / \ |atan2| + - || / / \ / \ |atan2| + - ||
n5 = - 4 / | + | + | - | *cos|| - i*4 / | + | + | - | *sin||
n6 = 4 / | + | + | - | *cos|| + i*4 / | + | + | - | *sin||
(-1; 2) , (2; - 1).
Объяснение:
1) {х³ + у³ = 7
{ху(х+у) = - 2;
{(х+ у)(х²-ху+у²) = 7
{(х+ у)((х+у)² -3ху) = 7
Пусть х+у = а; xy = b, получим
{а(а² - 3b) = 7,
{ba = - 2;
{а³ - 3ba = 7,
{а³ + 6 = 7,
{а³ = 1,
{а = 1,
{a = 1,
{b = - 2.
2) Получили, что
{х + у = 1,
{ху = - 2.
{х = 1 - у
{(1-у)у = - 2
{-у² + у = - 2
{у² - у - 2 = 0;
{ х = 1 - у,
{ у = 2 или у = - 1
{х = - 1. или {х = 2
{у = 2; {у = - 1.
(-1; 2) , (2; - 1)
Проверка:
1) (-1; 2)
{(-1)³ + 2³ = 7 - верно;
{ -2•(-1 + 2) = -2 - верно.
2) (2; - 1)
{2³ + (-1)³ = 7 - верно;
{ -2•(2-1) = -2 - верно
4 / 2 2 /atan2(-im(m), -re(m))\ 4 / 2 2 /atan2(-im(m), -re(m))\
\/ 3 *\/ im (m) + re (m) *cos|| i*\/ 3 *\/ im (m) + re (m) *sin||
\ 2 / \ 2 /
n1 = - -
3 3
4 / 2 2 /atan2(-im(m), -re(m))\ 4 / 2 2 /atan2(-im(m), -re(m))\
\/ 3 *\/ im (m) + re (m) *cos|| i*\/ 3 *\/ im (m) + re (m) *sin||
\ 2 / \ 2 /
n2 = +
3 3
/ / \\ / / \\
/ 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)|| / 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)||
/ / \ / \ |atan2| - + || / / \ / \ |atan2| - + ||
/ |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /| / |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /|
n3 = - 4 / | - | + | + | *cos|| - i*4 / | - | + | + | *sin||
\/ \ 6 6 / \ 6 6 / \ 2 / \/ \ 6 6 / \ 6 6 / \ 2 /
/ / \\ / / \\
/ 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)|| / 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)||
/ / \ / \ |atan2| - + || / / \ / \ |atan2| - + ||
/ |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /| / |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /|
n4 = 4 / | - | + | + | *cos|| + i*4 / | - | + | + | *sin||
\/ \ 6 6 / \ 6 6 / \ 2 / \/ \ 6 6 / \ 6 6 / \ 2 /
/ / \\ / / \\
/ 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)|| / 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)||
/ / \ / \ |atan2| + - || / / \ / \ |atan2| + - ||
/ |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /| / |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /|
n5 = - 4 / | + | + | - | *cos|| - i*4 / | + | + | - | *sin||
\/ \ 6 6 / \ 6 6 / \ 2 / \/ \ 6 6 / \ 6 6 / \ 2 /
/ / \\ / / \\
/ 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)|| / 2 2 | |im(m) \/ 3 *re(m) re(m) \/ 3 *im(m)||
/ / \ / \ |atan2| + - || / / \ / \ |atan2| + - ||
/ |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /| / |im(m) \/ 3 *re(m)| |re(m) \/ 3 *im(m)| | \ 6 6 6 6 /|
n6 = 4 / | + | + | - | *cos|| + i*4 / | + | + | - | *sin||
\/ \ 6 6 / \ 6 6 / \ 2 / \/ \ 6 6 / \ 6 6 / \ 2 /
(-1; 2) , (2; - 1).
Объяснение:
1) {х³ + у³ = 7
{ху(х+у) = - 2;
{(х+ у)(х²-ху+у²) = 7
{ху(х+у) = - 2;
{(х+ у)((х+у)² -3ху) = 7
{ху(х+у) = - 2;
Пусть х+у = а; xy = b, получим
{а(а² - 3b) = 7,
{ba = - 2;
{а³ - 3ba = 7,
{ba = - 2;
{а³ + 6 = 7,
{ba = - 2;
{а³ = 1,
{ba = - 2;
{а = 1,
{ba = - 2;
{a = 1,
{b = - 2.
2) Получили, что
{х + у = 1,
{ху = - 2.
{х = 1 - у
{(1-у)у = - 2
{х = 1 - у
{-у² + у = - 2
{х = 1 - у
{у² - у - 2 = 0;
{ х = 1 - у,
{ у = 2 или у = - 1
{х = - 1. или {х = 2
{у = 2; {у = - 1.
(-1; 2) , (2; - 1)
Проверка:
1) (-1; 2)
{(-1)³ + 2³ = 7 - верно;
{ -2•(-1 + 2) = -2 - верно.
2) (2; - 1)
{2³ + (-1)³ = 7 - верно;
{ -2•(2-1) = -2 - верно