a) 1/7 -(-3/14) +2                         b) 2/5 + 1/5 .(-3/4)               c) -3,75.(-7,2) + 2,8.3,75                      d) 1:( 2/3 +3/4 ) ^ 2 ( ^ dấu mũ nha )

= 1/7 + 3/14 +2                            = 2/5 + (-3/20)                      = 3,75 .7,2 + 2,8 . 3,75                        = 1 : (8/12 +9/12) ^ 2

= 2/14 + 3/14 + 7/14                    = 8/20 + (-3/20)                    = 3,75.(7,2+2,8)                                   = 1: (17/12) ^2 

= 12/14 = 6/7                               = 5/20 = 1/4                          = 3,75.1= 3,75                                     = 1.(17/12) ^2 = 289/144

A = \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}.\left(\sqrt{5}+\sqrt{27}\right)}\)

A = \(\frac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}.\left(\sqrt{3}-2\sqrt{2}\right)}-\frac{1}{\sqrt{6}}=-\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)

b) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)

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

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

\(A=4.2.10=80\)

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

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