$\begin{cases}log_2(x^2+1)+log_\frac{1}{2}2(y^2+1)=-1 \\ 4^{2x+y}+5.2^{x+2y}=6 \end{cases}$
$\begin{cases}\log_2(x^2+1)+ \log_\frac{1}{2}2(y^2+1)=-1(1)\\ 4^{2x+y}+5.2^{x+2y}=6(2) \end{cases}$
$(1)\Leftrightarrow \log _2(x^2+1)-\log _22(y^2+1)=-1$
        $\Leftrightarrow \log _2\frac{x^2+1}{2(y^2+1)}=-1\Leftrightarrow x^2=y^2\Leftrightarrow x=\pm y$
Với $x=y$, ta có:
$(2)\Leftrightarrow 4^{3x}+5.2^{3x}-6=0\Leftrightarrow 2^{3x}=1\Leftrightarrow x=0\Rightarrow y=0.$
Với $x=-y$, ta có:
$(2)\Leftrightarrow 4^x+5.2^{-x}-6=0\Leftrightarrow (2^{x})^3-6.2^x+5=0$
        $\Leftrightarrow 2^x=1 $ hoặc $2^x=\frac{-1+\sqrt{21}}{2}$
        $\Leftrightarrow x=0\Rightarrow y=0$ hoặc $x=\log _2\frac{-1+\sqrt{21}}{2}\Rightarrow y=-\log _2\frac{-1+\sqrt{21}}{2}.$
Kết luận:$..........................$

Cần trả +10,000vỏ sò để xem nội dung lời giải này

$(1)\Leftrightarrow \log_2(x^2+1)-(\log_22+\log_2(y^2+1))=-1 $
$\Leftrightarrow \log_2(x^2+1)-\log_2(y^2+1)=0$
$\Leftrightarrow \log_2\frac{x^2+1}{y^2+1}=0\Leftrightarrow x^2+1=y^2+1\Leftrightarrow x=\pm y$
Với $x=y$
$(2)\Leftrightarrow 4^{3x}+5.2^{3x}-6=0$
Đặt $t=2^{3x}>0$
$(2)\Rightarrow t^2+5t-6=0\Leftrightarrow t=1\Leftrightarrow 2^{3x}=1\Leftrightarrow x=0$
Với $x=-y$
Tương tự ta có: $(t-1)(5t^2-t-1)=0\Leftrightarrow t=1\vee t=\frac{1+\sqrt{21}}{10}$ với $t=2^x$
$\Leftrightarrow x=y=0\vee x=\log_2\frac{1+\sqrt{21}}{10}\Rightarrow y=-\log_2\frac{1+\sqrt{21}}{10}$
Vậy nghiệm hệ $(x;y)=(0;0),(\log_2(\frac{1+\sqrt{21}}{10};-\log_2\frac{1+\sqrt{21}}{10})$

cham qua ha –  ★★.P.I.N.O.★★ 26-06-14 10:38 PM
khổ thế, đang đăng bài bên kia ko có thời gian –  Nero 26-06-14 10:37 PM
x=-y co 1 nghiem khac nua –  ★★.P.I.N.O.★★ 26-06-14 10:25 PM
Thiếu trường hợp: x=y vs x=-y –  ★★.P.I.N.O.★★ 26-06-14 09:59 PM
Thấy đúng thì nhấn vào biểu tượng chữ V màu trắng bên dưới vote down nhé! Lần sau mình sẽ ss giúp đỡ. Tks bạn –  Nero 26-06-14 09:50 PM

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