A=1+3+3²+3³+......+3²⁰⁰⁸

3²A=3²+3³+3⁴+3⁵+......+3²⁰¹⁰

9A-A=(3²+3³+3⁴+3⁵+......+3²⁰¹⁰)-(1+3+3²+3³+.....+3²⁰⁰⁸)

8A=3²⁰¹⁰-(1+3)

8A=3²⁰¹⁰-4

B=8A-3²⁰¹⁰

-->B=3²⁰¹⁰-4-3²⁰¹⁰

B= -4

Vậy B= -4

\(A=1+3^2+3^4+...+3^{2008}\)

\(9A=3^2+3^4+...+3^{2008}+3^{2010}\)

\(\Rightarrow8A=3^{2010}-1\)

\(\Rightarrow B=3^{2010}-1-3^{2010}=-1\)

a thuộc  [1;1] là sao bạn? 

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)

\(\rightarrow a=2k;b=3k;c=4k\)

\(M=\dfrac{3a+2b-4c}{8a-5b+2c}\\ =\dfrac{3.2k+2.3k-4.4k}{8.2k-5.3k+2.4k}\\ =\dfrac{6k+6k-8k}{16k-15k+8k}\\ =\dfrac{4k}{9k}=\dfrac{4}{9}\)

Vậy \(M=\dfrac{4}{9}\)

nhưng phải biết giá trị của A rồi mới tính đc chứ

giải

A = 3+32+33+34+35+36+37+38+...+32010+32011+32012

A = (3+32+33+34)+(35+36+37+38)+...+(32009+32010+32011+32012)

A = 120+34.120+...+32008.120

A = 120.(1+34+...+32008) ⋮120

VẬY A chia hết cho120 (ĐPCM)

Ta có :

A = 1 + 3² + 3⁴ + 3⁶ + .... + 3²⁰⁰⁸.

=> 9A = 3² + 3⁴ + 3⁶ + 3⁸ + .... +3²⁰¹⁰

=> 9A - A = ( 3² + 3⁴ + 3⁶ + 3⁸ + .... + 3²⁰¹⁰) - (1 + 3² + 3⁴ + 3⁶ + .... + 3²⁰⁰⁸)

=> 8A = -1 + 3²⁰¹⁰

=> 8A - 3²⁰¹⁰ = -1

THANKS CÁC BẠN NHA !!!!! 

\(P=\frac{a+2}{a+3}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{a}{a^2-2a}\)

a) ĐKXĐ : \(\hept{\begin{cases}a\ne0\\a\ne-3\\a\ne2\end{cases}}\)

b)\(=\frac{a+2}{a+3}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{a}{a\left(a-2\right)}\)

\(=\frac{a\left(a-2\right)\left(a+2\right)}{a\left(a+3\right)\left(a-2\right)}-\frac{5a}{a\left(a+3\right)\left(a-2\right)}-\frac{a\left(a+3\right)}{a\left(a+3\right)\left(a-2\right)}\)

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

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

\(=\frac{a^3-a^2-12a}{a\left(a+3\right)\left(a-2\right)}=\frac{a\left(a^2-a-12\right)}{a\left(a+3\right)\left(a-2\right)}\)

\(=\frac{a^2-4a+3a-12}{\left(a+3\right)\left(a-2\right)}=\frac{a\left(a-4\right)+3\left(a-4\right)}{\left(a+3\right)\left(a-2\right)}\)

\(=\frac{\left(a-4\right)\left(a+3\right)}{\left(a+3\right)\left(a-2\right)}=\frac{a-4}{a-2}\)

c) \(8a=8a^2\)

⇔ \(8a^2-8a=0\)

⇔ \(8a\left(a-1\right)=0\)

⇔ \(\orbr{\begin{cases}8a=0\\a-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=0\left(ktm\right)\\a=1\left(tm\right)\end{cases}}\)

Với a = 1 =>\(P=\frac{1-4}{1-2}=\frac{-3}{-1}=3\)

\(a+b=6\)

<=>   \(\left(a+b\right)^2=36\)

<=>   \(a^2+2ab+b^2=36\)

<=>   \(2ab=36-a^2-b^2=-1974\)

<=>   \(ab=--987\)

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=6^3-3.\left(-987\right).6=17982\)

\(a^3+b^3=\left(a+b\right)\left(a^2+2ab+b^2\right)\)

\(=6\left(2010+2ab\right)\)

\(12060+6\left[\left(a+b\right)^2-a^2-b^2\right]\)

\(12060+6\left(36-2010\right)\)

\(=12060-11844\)

\(=216\)

sửa lại 

\(2ab=\left(a+b\right)^2-a^2-b^2\)

\(2ab=36-2010\)

\(2ab=-1974\)

\(\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(6\left(2010-ab\right)\)

\(12060-\left(-5922\right)\)

\(=17892\)

Bài 1:

a) Số nguyên dương nhỏ nhất là 1

Do đó, ta có : x + 2011 = 1

x = 1 – 2011 = -2010

b) Các số nguyên có giá trị tuyệt đối nhỏ hơn 100 là -99 ; -98 ; … ; 98 ; 99

Tổng cần tìm là: ( -99 + 99 ) + ( -98 + 98 ) + … + ( -1 + 1 ) + 0 = 0 + 0 + ... + 0 = 0