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Bài 1:
Đk:\(1\le x\le2\)
\(pt\Leftrightarrow x^2-3x-10=-\sqrt{x^2-3x+2}\)
Đặt \(\sqrt{x^2-3x+2}=t\left(t\ge0\right)\) ta có:
\(t^2-12=-t\Leftrightarrow t^2+t-12=0\)\(\Leftrightarrow\left(t+4\right)\left(t-3\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-4\left(loai\right)\\t=3\end{array}\right.\)
Xét \(t=3\Leftrightarrow x^2-3x+2=3\)
\(\Leftrightarrow x^2-3x-1=0\)
\(\Delta=\left(-3\right)^2-\left[\left(-4\right).\left(1.1\right)\right]=13\)\(\Leftrightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\) (thỏa mãn)
Cho sin a=
pt(1)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\3x^2+\left(6+y^2\right)x+2y^2=0\left(1'\right)\end{array}\right.\)
*)x=0.Thay vào pt(2) ta đc:y\(^2\)=-3(VN)
*)(1')\(\Leftrightarrow\left(x+2\right)\left(y^2+3x\right)=0\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\y^2=-3x\end{array}\right.\)
TH1:x=-2\(\Rightarrow y^2\)=-5(VN)
TH2:y\(^2\)=-3x.(x\(\le0\)).Thay vào pt(2) ta đc:\(^2\)x\(^2\)
\(\Rightarrow\)x=3(L) hoặc x=1(L)
Vậy hệ pt vô nghiệm
\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)
b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)
c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)
d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)
\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)
e/