a) tan 19 

cot 40 = tan 50 

Vì 19 < 50 

Nên tan 19 < tan 50 

Vậy tan 19 < cot 40 

b) sin 36 

cos 71 = sin 19 

Vì 36 > 19 

Nên sin 36 > sin 19 

Vậy sin 36 > cos 71

7^3 :7 -7^2
= 7^3:7^1-7^2
=7^(3-1)-7^2
=7^2-7^2
=0

nhầm ạ mn thông cảm

Ta có : F + G = 90° 

=》G = 90° - 60° = 30°

sinF = EG/FG 

=》 sin60° = EG/8 

=> EG = 8 x sin60°

=》EG \(\approx\)6,9282 (cm) 

sinG = EF/FG 

=》 sin30° = EF/8

=> EF = 8 x sin30°

=》 EF = 4 (cm) 

=\(\sqrt{15-6\sqrt{10}+6}\) 

=\(\sqrt{\left(\sqrt{15}\right)^2+2\cdot\sqrt{15}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)  

=\(\sqrt{\left(\sqrt{15}+\sqrt{6}\right)^2}\) 

=\(|\sqrt{15}+\sqrt{6}|\) 

=\(\sqrt{15}+\sqrt{6}\) 

=\(\sqrt{3}\left(\sqrt{5}+\sqrt{2}\right)\)

\(\sqrt{21-6\sqrt{10}}\)

\(=\sqrt{15-6\sqrt{10}+6}\)

\(=\sqrt{\left(\sqrt{15}\right)^2-2\cdot\sqrt{15}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)

\(=\sqrt{\left(\sqrt{15}-\sqrt{6}\right)^2}\)

\(=\left|\sqrt{15}-\sqrt{6}\right|\)

\(=\sqrt{15}-\sqrt{6}\)

\(=\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)\)

a) \(A=\sqrt{4x^2+4x+2}=\sqrt{4x^2+4x+1+1}=\sqrt{\left(2x+1\right)^2+1}\)

Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)

\(\Rightarrow A\ge\sqrt{1}=1\)

Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)

b) \(B=\sqrt{2x^2-4x+5+1}=\sqrt{2x^2-4x+2+3+1}=\sqrt{2\left(x^2-2x+1\right)+4}\)

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

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow B\ge\sqrt{4}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(minB=2\Leftrightarrow x=1\)

Mơn bạn nha

rút gọn ạ

\(\frac{\sqrt{10}+\sqrt{5}}{\sqrt{2}+1}+\frac{3\sqrt{5}-5}{\sqrt{3}-3}-\frac{3}{\sqrt{3}}\)

\(=\sqrt{5}+\frac{3\sqrt{5}-5}{\sqrt{3}-3}-\sqrt{3}\)

\(\approx2.620900819\) (nguồn : kết quả từ máy tính \(fx-570VN\) \(PLUS\)

k mình nha

a. Không giải được\(\sqrt{29}-6\sqrt{6}< 0\)     

b. \(\left(\sqrt{8}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(2\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

a) Không thể giải vì \(\sqrt{29}-6\sqrt{6}< 0\) 

b) \(\left(\sqrt{8}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(2\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(-\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(-2-2\sqrt{5}-2\sqrt{5}\) 

=\(-2-4\sqrt{5}\) 

=\(-2\left(1+2\sqrt{5}\right)\)

\(=\sqrt{4+2-4\sqrt{2}}\)

\(=\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

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

\(\sqrt{6-4\sqrt{2}}\)

\(=\sqrt{2-4\sqrt{2}+4}\)

\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot2\cdot\sqrt{2}+2^2}\)

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

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

\(=-\left(\sqrt{2}-2\right)=2-\sqrt{2}\)( vì \(\sqrt{2}< 2\))