a)M=2¹⁰⁰-2⁹⁹+2⁹⁸-...+2²-2

2²M=2¹⁰²-2¹⁰¹+2¹⁰⁰-...+2^2-2

4M-M=2¹⁰²-2¹⁰¹+2¹⁰⁰-...+2²-2-2¹⁰⁰+2⁹⁹-...-2²+2

3M=2¹⁰²-2¹⁰¹

M=\(\frac{2^{102}-2^{101}}{3}\)

chết lm sai mất òy

\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)

\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)

a) 5A = 5 + 5^2 + 5^3 + 5^4 +...+ 5^51

=> 5A - A = 4A = 5^51 - 1

=> A = \(\frac{5^{51}-1}{4}\)

b) 3B = 3^100 - 3^99 -...- 3

=> 3B - B = 2B = 3^100 - 2.3^99 + 1

=> B = \(\frac{3^{100}-2\times3^{99}+1}{2}\)

a, 1+5+5²+.....+5⁵⁰

=> 5(1+5+5²+.....+5⁵⁰)=5+5²+5³.....+5⁵¹

=>4(1+5+5²+.....+5⁵⁰)=5⁵¹-1

=>1+5+5²+.....+5⁵⁰=(5⁵¹-1):4

b,3⁹⁹-3⁹⁸-...-3-1

=3⁹⁹-(3⁹⁸+...+3+1)

=3⁹⁹-(3⁹⁹-1):2

Làm như bạn Linh nguyen phan khanh có nghĩa là chỉ có 3^99 - 1 chia 2 thôi à, nếu thế thì sai rồi nhé

a: A=-a+b-c+a+b+c

=2b

b: Khi a=1; b=-1; c=-2 thì A=2*(-1)=-2

Thay dấu * bằng chữ số thích hợp để mỗi số sau là số nguyên tố:
a) 3*; b) *1 ; c) 1*5 .

a: \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{2\left(x-3\right)}{2-x}\)

\(=\dfrac{4+4x+x^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{2\left(x-3\right)}\)

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

\(=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{1}{2\left(x-3\right)}=\dfrac{4x\left(x+2\right)}{2\left(x+2\right)}\cdot\dfrac{1}{x-3}=\dfrac{2x}{x-3}\)

b: |x-2|=2

=>x-2=2 hoặc x-2=-2

=>x=0(nhận) hoặc x=4(nhận)

Khi x=0 thì \(A=\dfrac{2\cdot0}{0-3}=\dfrac{-2}{3}\)

Khi x=4 thì \(A=\dfrac{2\cdot4}{4-3}=8\)

c: A>0

=>x/x-3>0

=>x>3 hoặc x<0

=>x>3

 a) A= -a+b-c + a+b+c = 2b 
b) Vì giá trị của A = không phụ thuộc vào a hay c nên A=2b=2.(-1)= -2

a, Rút gọn 

A = ( - a - b - c ) - ( - a - b - c )

   = - a - b - c - a - b - c 

  = 2b

A=-a-b-c+a+b+c

A=0

a) 

\(A=\frac{6^3+3.6^3+3^3}{-13}=\frac{3^3.2^3+3^3.2^2+3^3}{-13}=\frac{3^3\left(8+4+1\right)}{-13}=\frac{27.13}{-13}=-27\)

b)

A=1+5+5²+5³+...+5⁵⁰

5A=5+5²+5³+...5⁵¹

5A-A=(5+5²+5³+...+5⁵¹)-(1+5+5²+...+5⁵⁰)

4A=5⁵¹-1

A=\(\frac{5^{51}-1}{4}\)

c)

A=2¹⁰⁰-2⁹⁹+2⁹⁸-...+2²-2

2A=2¹⁰¹-2¹⁰⁰+2⁹⁹-...+2³-2²

2A+A=(2¹⁰¹-2¹⁰⁰+...+2³-2²)+(2¹⁰⁰-2⁹⁹+...+2²-2)

3A=2¹⁰¹-2

A=\(\frac{2^{101}-2}{3}\)

b.

\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)

\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)

\(5A-A=\left(5+5^2+5^3+...+5^{50}+5^{51}\right)-\left(1+5+5^2+..+5^{50}\right)\)

\(4A=5^{51}-1\)

\(A=\frac{5^{51}-1}{4}\)

a) 

\(A=\frac{6^3+3.6^2+3^3}{13}=\frac{2^3.3^3+3.3^2.2^2+3^3}{13}=\frac{3^3\left(2^3+2^2+1\right)}{13}=\frac{27.13}{13}=27\)

b)

A=1+5+5²+5³+...+5⁵⁰

5A=5+5²+5³+...5⁵¹

5A-A=(5+5²+5³+...+5⁵¹)-(1+5+5²+...+5⁵⁰)

4A=5⁵¹-1

A=\(\frac{5^{51}-1}{4}\)

c)

A=2¹⁰⁰-2⁹⁹+2⁹⁸-...+2²-2

2A=2¹⁰¹-2¹⁰⁰+2⁹⁹-...+2³-2²

2A+A=(2¹⁰¹-2¹⁰⁰+...+2³-2²)+(2¹⁰⁰-2⁹⁹+...+2²-2)

3A=2¹⁰¹-2

A=\(\frac{2^{101}-2}{3}\)

Xét mẫu số của phân số:

\(\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}\)

\(=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+\left(\dfrac{99}{1}-98\right)\)

\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)

\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)

Ta thấy mẫu số gấp tử số 100 lần. Vậy phân số đó có giá trị bằng \(\dfrac{1}{100}\)