\(A=\sqrt{32.200}=\sqrt{32}\sqrt{200}=4\sqrt{2}.\sqrt{2}\sqrt{100}\)

\(A=4.2.10=80\)

\(B=\sqrt{5}.\sqrt{125}=\sqrt{5}.5\sqrt{5}=5.5=25\)

a) Đường thẳng (d) đi qua A(1; 0) => x = 1 và y = 0

DO đó: 0 = m - 3 <=> m = 3

b) pt hoành độ giao điểm giữa (P) và (d) là:

 x² = mx - 3 <=> x² - mx + 3 = 0 (1)

\(\Delta\)= (-m)² - 3.4 = m² - 12

Để (P) cắt (d) tại 2 điểm pb <=>  pt (1) có 2 nghiệm pb 

<=> \(\Delta\)> 0 <=> m² - 12 > 0 <=> \(\orbr{\begin{cases}m>2\sqrt{3}\\m< -2\sqrt{3}\end{cases}}\)

Theo hệ thức viet, ta có: \(\hept{\begin{cases}x_1+x_2=m\\x_1x_2=3\end{cases}}\)

Theo bài ra, ta có: |x₁ - x₂| = 2

<=> x₁² - 2x₁x₂ + x₂² = 4

<=> (x₁ + x₂)² - 4x₁x₂ = 4

<=> m² - 4.3 = 4

<=> m² - 16 = 0

<=> (m  - 4)(m + 4) = 0

<=> \(\orbr{\begin{cases}m=4\\m=-4\end{cases}}\)(tm)

662.724678882566

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