Question

Giải hệ phương trình :

          \(\begin{cases}\left(4x+2\right)^2=2y+15\left(1\right)\\\left(4y+2\right)^2=2x+15\left(2\right)\end{cases}\)

Answer 1

Trừ 2 vế ta được: (4x + 2)² - (4y + 2)² = 2y - 2x  => (4x + 2 + 4y + 2).(4x + 2 - 4y - 2) + 2x - 2y = 0

=> (4x + 4y + 4).(4x - 4y) + 2.(x - y) = 0

=> 16.(x + y + 1).(x - y) + 2.(x - y) = 0

=> 8.(x + y + 1).(x - y) + 2.(x - y) = 0

=> (x - y). (8x + 8y + 8 + 2) = 0

=> (x - y).(8x + 8y + 10) = 0

=> (x - y).(4x + 4y + 5) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x=y\\4x+4y+5=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=y\\x=\frac{-5-4y}{4}\end{array}\right.\)

Tới đây bạn chia ra 2 trường hợp giải nha

Answer 2

Lấy (2) trừ (1), ta có :

\(\left(4x-4y\right)\left(4x+4y+4\right)=2y-2x\)

\(\Leftrightarrow2\left(x-y\right)\left(8x+8y+9\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-y=0\\8x+8y+9=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}y=x\\y=-\frac{8x+9}{8}\end{array}\right.\)

* Với \(y=x\), thay vào (1) ta có :

\(\left(4x+2\right)^2=2x+15\)

\(\Leftrightarrow16x^2+14x-11=0\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{11}{8}\end{array}\right.\)

Vậy \(\left(x;y\right)=\left(\frac{1}{2};\frac{1}{2}\right);\left(x;y\right)=\left(-\frac{11}{8};-\frac{11}{8}\right)\) là nghiệm của hệ phương trình 

* Với \(y=-\frac{8x+9}{8}\), ta có : 

\(\left(4x+2\right)^2=15-\frac{8x+9}{4}\)

\(\Leftrightarrow64x^2+72x-35=0\)

\(\Leftrightarrow x=\frac{-9\pm\sqrt{221}}{16}\)

Khi \(x=\frac{-9-\sqrt{221}}{16}\Rightarrow y=\frac{-9+\sqrt{221}}{16}\)

Khi \(x=\frac{-9+\sqrt{221}}{16};y=\frac{-9-\sqrt{221}}{16}\)

Hệ đã cho có 4 nghiệm :

\(\left(\frac{1}{2};\frac{1}{2}\right);\left(-\frac{11}{8};-\frac{11}{8}\right);\left(\frac{-9-\sqrt{221}}{16};\frac{-9+\sqrt{221}}{16}\right);\left(\frac{-9+\sqrt{221}}{16};\frac{-9-\sqrt{221}}{16}\right)\)