Решите систему уравнений
[tex] \left \{ {{(4x+3) ^{2}=7y } \atop {(3x+4) ^{2}=7y }} \right. [/tex]
[tex] \left \{ {{(4x+3) ^{2}=7y } \atop {(3x+4) ^{2}=7y }} \right. [/tex]
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Answers & Comments
(4х+3)²=(3х+4)²
16х²+24х+9=9х²+24х+16
16х²+24х+9-9х²-24х-16=0
7х²-7=0
7х²=7
х²=7/7=1
х₁=1; х₂=-1
7у=(4х+3)²=(4+3)²=49 ⇒ у=49/7=7
7у=(4х+3)²=(-4+3)²=1 ⇒ у=1/7
отв:(1;7), (-1;1/7)