Ta có:

sin 2 80 0 = cos 2 10 0 ;  sin 2 70 0 = cos 2 20 0

sin 2 60 0 = cos 2 30 0 ;  sin 2 50 0 = cos 2 40 0

Nên

sin 2 10 0  +  sin 2 20 0  +  sin 2 30 0  +  sin 2 40 0  +  sin 2 50 0  +  sin 2 60 0  +  sin 2 70 0  +  sin 2 80 0

=  sin 2 10 0  +  sin 2 20 0  +  sin 2 30 0  +  sin 2 40 0  +  cos 2 40 0 +  cos 2 30 0  +  cos 2 20 0  +  cos 2 10 0

= ( sin 2 10 0  +  cos 2 10 0 ) + ( sin 2 20 0  +  cos 2 20 0 ) + ( sin 2 30 0  +  cos 2 30 0 ) + ( sin 2 40 0  +  cos 2 40 0 )

= 1 + 1 + 1 + 1 = 4

Vậy giá trị cần tìm là 4

Đáp án cần chọn là: D

C=1.2+2.3+...+99.100

3C=1.2.3+2.3.3+...+99.100.3

3C=1.2(3-0)+2.3(4-1)+...+99.100(101-98)

C=99.100.101 phần 3

C=333 300

D=2²+4²+6²+333 300

D=4+16+36+333 300

D=20+36+333 300

D=56+333 300

D=333 356

Tích cho mình hai câu nhà, thank

a) \(C=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(\Rightarrow3C=99.100.101\)

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

a) C = 1.2 + 3.4 + 4.5 +...+ 99.100

\(\Rightarrow\) C = 1.2+2.3+3.4+4.5+...+99.100

\(\Rightarrow\)3C = 1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

\(\Rightarrow\)3C = 1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

 \(\Rightarrow\)3C = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

 \(\Rightarrow\)3C = 1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101 

\(\Rightarrow\)3C = 99.100.101

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

b) D = 2 + 4 + 6 + 333 300

\(\Rightarrow\)D = 4 + 16 + 36 + 333 300

\(\Rightarrow\)D = 20 + 36 + 333 300 

\(\Rightarrow\)D = 56 + 333 300

\(\Rightarrow\)D = 333 356 

a, C = 1.2 + 2.3 + 3.4 + ... + 99.100

3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

3C = 1.2.3 + 2.3\((4-1)\)+ 3.4\((5-2)+...\)+ 99.100\((101-98)\)

3C = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3C = 99.100.101

C = 99.100.101 :3

C = 333300

Câu b tự làm

P/S : Hoq chắc :>

a)đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

A=1.2+2.3+3.4+4.5+...+99.100

=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101 

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300 

b)D=2 +4 +6 +333 300

D=4+16+36+333 300

D=20+36+333 300 

D=56+333 300

D=333 356 

thank

để A xác định

\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)

\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)

\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)

\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)

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

C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)

d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)

\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)

\(x^2-4\inℕ\)

\(\Rightarrow x^2\in\left\{0;4;12\right\}\)

Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn

mk học lớp 6 mong mn thông cảm nếu có sai sót

Mk nhầm nha

=2/x-2 nha bạn

b, =1/-3

c, 2 chia hết cho x-1

=> x-1 E {-2;-1;1;2}

=> x E {-1;0;2;3} mk có tính nhầm 1 chỗ

a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)

b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)

Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)

c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)

Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)