giải pt:
1.   $1+sin^{3}4x+cos^{3}4x=3sin4xcos4x$
2.   $(sin^{2}x-6sinx-9)^{2}=sinx(sin^{2}x-4sinx-9)$
3.   $\left| {sin2x-cos2x} \right|+4sin4x=1$
3) $Pt\Leftrightarrow |sin2x-cos2x|+8sin2x.cos2x=1$
Đặt $t=sin2x-cos2x=\sqrt{2}(sin2x-\pi /4)$ $\Rightarrow sin2x.cos2x=\frac{1-t^2}{2}$
Thay vào pt ta có: $Pt\Leftrightarrow |t|+4(1-t^2)=1$ Bạn tự giải nốt pt chứa dấu giá trị tuyệt đối nhé
2) Xét $sinx=0$ Thay vào pt thấy không thỏa mãn (loại)
+) Xét $sinx\neq 0$ Chia 2 vế cho $sin^2x$ ta có: $Pt\Leftrightarrow (\frac{sin^2x-6sinx-9}{sinx})^2=\frac{sin^2x-4sinx-9}{sinx}$
$\Leftrightarrow (sinx-\frac{9}{sinx}-6)^2=(sinx-\frac{9}{sinx})-4$
Đến đây ta đặt    $sinx-\frac{9}{sinx}=t$
Ta có $Pt\Leftrightarrow (t-6)^2=t-4\Rightarrow ....................$ (bạn giải nốt nhé)
1)$Pt\Leftrightarrow 1+(sin4x+cos4x)(1-sin4x+cos4x)=3sin4x.cos4x$
Đặt $t=sin4x+cos4x=\sqrt{2}.sin(4x+\pi /4)$
Ta có: $sin4x.cos4x=\frac{t^2-1}{2}$
Thay vào pt ta có :$1+t(1-\frac{t^2-1}{2})=3(\frac{t^2-1}{2})$
Bạn tự giải nốt nhé

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