Câu hỏi của Ngô Chí Thành

Question

Rút gọn biểu thức $\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4sin^2x}$ .

Nguyễn Việt Lâm · Accepted Answer

$=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}$

$=\sqrt{4-4sin^2x+sin^4x}+\sqrt{4-4cos^2x+cos^4x}$

$=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}$

$=2-sin^2x+2-cos^2x=4-\left(sin^2x+cos^2x\right)$

$=3$Câu trả lời: $=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}$

$=\sqrt{4-4sin^2x+sin^4x}+\sqrt{4-4cos^2x+cos^4x}$

$=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}$

$=2-sin^2x+2-cos^2x=4-\left(sin^2x+cos^2x\right)$

$=3$