a) \(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{3-2\sqrt{15}+5}\)

\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)

\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(A=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)

\(A=2\left(16-15\right)=2\)

b) \(B=\left(3-\sqrt{15}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(B=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)

\(B=\sqrt{9-5}\cdot\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(B=2\cdot\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)\)

\(B=\sqrt{2}.2\sqrt{5}=2\sqrt{10}\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\sqrt{2+\sqrt{2+\sqrt{3}}}^2}\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\sqrt{2+\sqrt{3}}^2}\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)

đáp án

A=Sin 42^o - cos 48^o =cos(90^o - 42^o) - cos 48^o= cos48^o - cos48^o=0

hok tốt

B=cos56^o-tan34^o=tan(90^o - 56^o) - tan34^o=tan34^o - tan34^o=0

a) cos = 15/7

tan = 8/15

cot = 15/8

b) cos = 4/5

tan = 3/5

cot = 4/5

Xét tam giác ABC vuông tại A, đường cao AH

* Áp dụng hệ thức : \(AH^2=BH.CH=8.2=16\Rightarrow AH=4\)cm 

Áp dụng định lí Pytago tam giác ABH vuông tại H : 

\(AB^2=BH^2+AH^2=4+16=20\Rightarrow AB=2\sqrt{5}\)cm 

-> BC = BH + CH = 8 + 2 = 10 cm 

Áp dụng định lí Pytago tam giác ABC vuông tại A

\(BC^2=AB^2+AC^2\Rightarrow AC^2=BC^2-AB^2=100-20=80\Rightarrow AC=4\sqrt{5}\)cm 

* sinB = AC/BC = \(\frac{4\sqrt{5}}{10}=\frac{2\sqrt{5}}{5}\)

cosB = AB/BC = \(\frac{2\sqrt{5}}{10}=\frac{\sqrt{5}}{5}\)

tanB = AC/AB = \(\frac{4\sqrt{5}}{2\sqrt{5}}=2\)

cotaB = AB/AC \(\frac{2\sqrt{5}}{4\sqrt{5}}=\frac{1}{2}\)

Ta có: \(x^4-x^2-2mx-m^2=0\)

<=> \(x^4-\left(x+m\right)^2=0\)

<=> \(\left(x^2-x-m\right)\left(x^2+x+m\right)=0\)

<=> \(\orbr{\begin{cases}x^2-x-m=0\left(1\right)\\x^2+x+m=0\left(2\right)\end{cases}}\)

<=> \(\Delta_1=\left(-1\right)^2+4m=4m+1\)

 \(\Delta_2=1^2-4m=1-4m\)

Để pt có 4 nghiệm phân biệt <=> pt (1) và pt (2) cùng có 2 nghiệm pb

<=> \(\hept{\begin{cases}\Delta_1>0\\\Delta_2>0\end{cases}}\) <=> \(\hept{\begin{cases}4m+1>0\\1-4m>0\end{cases}}\) <=> \(\hept{\begin{cases}m>-\frac{1}{4}\\m< \frac{1}{4}\end{cases}}\) <=> \(-\frac{1}{4}< m< \frac{1}{4}\)

Vậy ...

\(\sqrt{18+6\sqrt{5}}+\sqrt{18-6\sqrt{5}}=\sqrt{\sqrt{15}^2+2\sqrt{45}+\sqrt{3}^2}+\sqrt{\sqrt{15}^2-2\sqrt{45}+\sqrt{3}^2}\)

\(=\sqrt{15}+\sqrt{3}+\sqrt{15}-\sqrt{3}\)

\(=2\sqrt{15}\)

2v15 nha