a, \(\cos^215+\cos^225+\cos^235+\cos^245+\sin^235+\sin^225+\sin^215\)

=\(\left(\cos^215+\sin^215\right)+\left(\cos^225+\sin^225\right)+\left(\cos^235+\sin^235\right)+\cos^245\)

=\(1+1+1+\frac{1}{2}=\frac{7}{2}\)

b.\(\sin^210-\sin^220-\sin^230-\sin^240-\cos^240-\cos^220+\cos^210\)

=\(\left(\sin^210+\cos^210\right)-\left(\sin^220+\cos^220\right)-\left(\sin^240+\cos^240\right)-\sin^230\)

=\(1-1-1-\frac{1}{4}=-\frac{5}{4}\)

c,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)

Trung bình cộng của 4 số đó là:

(220 + 240 + 260 + 280) : 4 = 250

Chọn A. 250

câu a "cot" chuyển thành "cos" giùm mjk nha

https://hoc24.vn/hoi-dap/question/647714.html

a) ta có : 

\(A=sin^210+sin^220+sin^230+sin^280+sin^270+sin^260=sin^210+sin^220+sin^230+cos^210+cos^220+cos^230=1+1+1=3\)\(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)+\left(1+cot^2\alpha\right)\left(1-cos^2\alpha\right)=\dfrac{1}{cos^2\alpha}.cos^2\alpha+\dfrac{1}{sin^2\alpha}.sin^2\alpha=1+1=2\)

B

Tổng của 1 tứ giác bằng :

A : 180 độ

B : 360 độ

C : 270 độ 

D : 260 độ

B

Bài 1 :

\(D=cos^220^0+cos^230^0+cos^240^0+cos^250^0+cos^260^0+cos^270^0\)

\(=\left(cos^220^0+cos^270^0\right)+\left(cos^230^0+cos^260^0\right)+\left(cos^240^0+cos^250^0\right)\)

\(=1+1+1=3\)

Bài 2 :

\(E=sin^25^0+sin^225^0+sin^245^0+sin^265^0+sin^285^0\)

\(=\left(sin^25^0+sin^285^0\right)+\left(sin^225^0+sin^265^0\right)+sin^245^0\)

\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)

Bài 3 :

\(F=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)

\(=1-3sin^2\alpha.cos^2\alpha+3sin^2a.cos^2\alpha\)

\(=1\)

Tính một cách hợp lí:

a) 2 834 + 275  - 2 833 - 265;

= (2 834 – 2 833) + (275 – 265)

= 1 + 10

= 11

b) (11 + 12 + 13) - (1 + 2 + 3).

= (11 – 1) + (12 – 2) + (13 – 3)

= 10 + 10 + 10

= 30

a) 2834 + 275 - 2833 - 265= 2833+1+275-2833-265=0+1+275-265=11

b) (11 + 12 + 13) - (1 + 2 + 3)=11 + 12 + 13-1-2-3=10+10+10=30

a) 2 834 + 275 - 2 833 - 265

=(275 - 265) + (2 834 - 2 833)

=10 + 1 = 11

b) (11 + 12 +13 ) - (1 + 2 + 3)

= 11 + 12 +13  - 1 - 2 - 3

= (11 - 1) + (12 - 2) +(13 - 3)

=10 +10 +10 =30

a) F = \(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}\right)+\frac{1}{2}.\left(\frac{1}{27}-\frac{1}{29}\right)+\frac{1}{2}.\left(\frac{1}{29}-\frac{1}{31}\right)+...+\frac{1}{2}.\left(\frac{1}{73}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\frac{2}{75}\)

F = \(\frac{1}{75}\)

b) G = \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)

G = \(\frac{15}{4}.\frac{4}{90.94}+\frac{15}{4}.\frac{4}{94.98}+\frac{15}{4}.\frac{4}{98.102}+...+\frac{15}{4}.\frac{4}{146.150}\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}\right)+\frac{15}{4}.\left(\frac{1}{94}-\frac{1}{98}\right)+\frac{15}{4}.\left(\frac{1}{98}-\frac{1}{102}\right)+...+\frac{15}{4}.\left(\frac{1}{146}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\frac{1}{225}\)

G = \(\frac{1}{60}\)

<br class="Apple-interchange-newline"><div id="inner-editor"></div>12.4 +14.6 +...+198.100 

=12 (22.4 +24.6 +...+298.100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −1100 )=12 .49100 =49200 

1056 +10140 +10260 +...+101400 =53 (

c) H = \(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

H = \(\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

H = \(\frac{5}{3}.\frac{3}{28}+\frac{5}{3}.\frac{3}{70}+\frac{5}{3}.\frac{3}{130}+...+\frac{5}{3}.\frac{3}{700}\)

H = \(\frac{5}{3}.\left(\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+...+\frac{3}{700}\right)\)

H = \(\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

H = \(\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

H = \(\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)

H = \(\frac{5}{3}.\frac{3}{14}\)

H = \(\frac{5}{14}\)

1)

a) -(2+5) = -2 - 5 = -7

b) +(-3+6) = -3 + 6 = 3

c) (-50+3) = -50 + 3 = -47

d) -(-2+3) = 2 - 3 = -1

e) -(10-3) = -10 + 3 = -7

f) -(-3)-(-3+1) = 3 + 3 - 1 = 5

g) (-5)+(-2+10) = -5 - 2 + 10 = 3

2)

a) -50+120+(-150)-20+30

= -(50 + 20) + (120 + 30 - 150)

= -70

b) 265-70+(-65)-30+15

= (265 - 65) - (70 + 30) + 15

= 200 - 100 + 15 = 115

c) -17+185-183+(-85)-63

= (185 - 85) - (183 + 17) - 63

= 100 - 200 - 63 = -163

d) -30+60+(-170)-260+19

= -(170 + 30) - (260 - 60) + 19

= -200 - 200 + 19 = -381