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hừm.................................

!Á

\(A^2=\left(\sin\alpha+\cos\alpha\right)^2\le2\left(sin^2\alpha+cos^2\alpha\right)=2\)

\(\Leftrightarrow A\le\sqrt{2}\)dấu bằng xảy ra khi \(\sin\alpha=\cos\alpha\)

\(B=\frac{1}{\sin^2\alpha}+\frac{1}{\cos^2\alpha}\ge\frac{4}{sin^2\alpha+cos^2\alpha}=4\)

dấu bằng xảy ra khi \(sin^2\alpha=cos^2\alpha\)

\(D=\left(\frac{3}{4}x+\frac{3}{x}\right)+\left(\frac{1}{4}x+\frac{1}{2}y\right)+\left(\frac{1}{2}y+\frac{9}{2y}\right)\ge3+\frac{1}{4}\left(x+2y\right)+3\ge3+2+3=8\)

dấu"=" xảy ra khi x=2;y=3

dat A =\(\sqrt{2+\sqrt{3}}\)+ \(\sqrt{2-\sqrt{3}}\)

=>A\(\sqrt{2}\)=\(\sqrt{4+2\sqrt{3}}\)+\(\sqrt{4-2\sqrt{3}}\)

=>A\(\sqrt{2}\)=\(\sqrt{3+2\sqrt{3}.1+1}\)+\(\sqrt{3-2\sqrt{3}.1+1}\)

=>A\(\sqrt{2}\)=\(\sqrt{\left(\sqrt{3}+1\right)^2}\)+\(\sqrt{\left(\sqrt{3}-1\right)^2}\)

=>A\(\sqrt{2}\)=\(\sqrt{3}\)+1+\(\sqrt{3}\)-1

=A\(\sqrt{2}\)=2\(\sqrt{3}\)

=>A=\(\frac{2\sqrt{3}}{\sqrt{2}}\)=\(\sqrt{2}\).\(\sqrt{3}\)=\(\sqrt{6}\)

cau b bn binh phuong len roi tinh nhe

1)x^4+x^2-6x+1=0>>>x^4+4x^2+4-3x^2-6x-3=0>>>(x^2+2)^2=3(x-1)^2.

>>Sau đó giải bt.

2)Đặt x^2-x+1=a;x+1=b thì:x^3+1=ab.

Pt:2a+5b^2+14ab=0(tự giải nha)