\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{2}.\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\text{|}\sqrt{5}-\sqrt{3}\text{|}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right).2=\left(16-15\right).2=2\)

Cảm ơn bạn nhiều nha!:))))

Sửa đề

\(\frac{sin^2x-c\text{os}^2x+c\text{os}^4x}{c\text{os}^2x-sin^2x+sin^4x}=\frac{sin^2x-c\text{os}^2x+\left(1-sin^2x\right)^2}{c\text{os}^2x-sin^2x+\left(1-c\text{os}^2x\right)^2}\)

\(=\frac{-sin^2x-c\text{os}^2x+sin^4x+1}{-c\text{os}^2x-sin^2x+c\text{os}^4x+1}\)

\(=\frac{-1+sin^4x+1}{-1+c\text{os}^4x+1}=\frac{sin^4x}{c\text{os}^4x}=tan^4x\)

$\sqrt{2+\sqrt{x}}$=3

<=> 2+√x=9

<=> \(\sqrt{x}\)=7

<=> x=49

Còn câu nữa mà???

khó quá

khó nhi

\(=\sqrt{\left(-2\right)^2}=\left|-2\right|=2\)

Làm hộ 1 cái thôi , mấy cái kia làm y hệt

\(1,x^2-2\left(m-1\right)x-3-m=0\)

Có: \(\Delta'=\left(m-1\right)^2+3+m\)

            \(=m^2-2m+1+3+m\)

            \(=m^2-m+4\)

             \(=\left(m-\frac{1}{2}\right)^2+\frac{15}{4}>0\forall m\)         

=> Pt luôn có nghiệm vs mọi m