a) 35-{12-[-14+(-2)]}.

= 35 - 12 + 14 + 2

= 35 + 14 - (12 - 2)

= 49 - 10

= 39

b) 160:{17+[9×5-(14+2048:256)]}.

= 160: {17 + [45 - (14 + 8)]}

= 160: {17 + [45 - 22)]}

= 160:{17 + 23}

= 160: 40 = 4

c) 2021⁰-{225:[20×15-8×25]-25}.

= 1 - {225:[300 - 200] - 25}

= 1 - {225 : 100 - 25}

= 1 - {2,25 - 25}

= 1 + 25 - 2,25

= 23,75

57713

a)\(\sqrt{\left(\frac{15}{16}\right)^2}=\frac{15}{16}\)

b)\(\sqrt{\frac{196}{10000}}=\sqrt{\frac{14^2}{100^2}}=\frac{14}{100}\)

Tính:

a,  . +  : 

 7 . 12 + 16 : 8 

   = 84 + 2

    = 86

b, 72 :  - 

 2.6.3-15

= -13

\(a,\sqrt{49}.\sqrt{144}+\sqrt{256}:\sqrt{64}\\ =7.12+16:8\\ =84+2\\ =86\\ b,72:\sqrt{2^3.3^2.36}-\sqrt{225}\\ =72:\sqrt{1296}-25\\ =72:36-25\\ =2-25\\ =-23\)

a)  ( − 225 ) . ( − 336 ) > 0

b)  ( − 112 ) .2. ( − 26 ) . ( − 321 ) < 0

Câu 34: B

Câu 35: B

Câu 34: B

Câu 35: B

sẽ bằng 2560

256*0+10*256 

=0+2560

=2560 

256 x 0 + 10 x 256 = 2560

a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)

\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

câu 1=0;câu 2=-123456789