a)Đặt \(L=\frac{1}{2^{2015}}+\frac{1}{2^{2014}}+...+\frac{1}{2^0}\)

\(2L=\left(1+\frac{1}{2}+...+\frac{1}{2^{2015}}\right)\)

\(2L=2+1+...+\frac{1}{2^{2014}}\)

\(2L-L=\left(2+1+...+\frac{1}{2^{2014}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2015}}\right)\)

\(2L=2-\frac{1}{2^{2015}}\) thay vào ta có:

\(B=\frac{1}{2^{2016}}-\left(2-\frac{1}{2^{2015}}\right)=\frac{1}{2^{2016}}-2+\frac{1}{2^{2015}}\)

b)Ta có:\(\begin{cases}\left|x+1\right|\ge0\\\left|x+4\right|\ge0\end{cases}\)\(\Rightarrow\left|x+1\right|+\left|x+4\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow3x\ge0\Rightarrow x\ge0\)

\(x+1+x+4=3x\)

\(\Rightarrow2x+5=3x\Rightarrow x=5\) (thỏa mãn)

Vậy x=5

â hình như sai

a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)

\(=\dfrac{1}{2016}\)

b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)

\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)

hay x=10

Vậy: x=10

Ta có : (6 - x)²⁰¹⁴ = (6 - x)²⁰¹⁵

=> (6 - x)²⁰¹⁴ - (6 - x)²⁰¹⁵ = 0

<=> (6 - x)²⁰¹⁴(1 - 6 - x) = 0

<=> \(\orbr{\begin{cases}\left(6-x\right)^{2014}=0\\1-6-x=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}6-x=0\\-5-x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=-5\end{cases}}\)

sory bạn trừng hợp hai mk nhầm : 

 1 - (6 - x) = 0

=> 1 - 6 + x = 0

=> -5 + x = 0

=> x = 5 

Ta có : 2 + 4 + 6 + ...... + 2x = 210

Nên : số số hạng là ; (2x - 2) : 2 + 1 = x - 1 + 1 = x (số)

Ta có : tổng là : (2x + 2) * x : 2 = 210

=> (2x² + 2x) : 2 = 210

=> x² + x = 210

=> x(x + 1) = 20.21

=> x = 20

a. \(\left[\left(-2\right)^5.2014-4^2.2015\right]-\left(-2015^0+3^2-2^3\right)\)

\(=-64448-32240+1-9+8=-96688\)

Tớ lm lại nhé:

SBC = 9-1/2-1/3-1/4-...-1/10

=1+1+...+1(9 số 1) -1/2-1/3-1/4-1/5-...-1/10.

=(1-1/2)+(1-1/3)+...+(1-1/10)

=1/2+2/3+...+9/10= SC

=> phép chia có thương là 1(vì SBC=SC)

SBC = (1-1/2) + (1-1/3) + ... + (1-1/10)

=1/2 + 2/3 +3/4 +... +9/10 = SC

Vậy thương là 1.

2A=2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ...+2/2014.2015.2016

Ta có: 2/1.2.3=1/1.2-1/2.3; 2/2.3.4=1/2.3-1/3.4; 2/3.4.5=1/3.4-1/4.5; ....; 2/2014.2015.2016=1/2014.2015-1/2015.2016

=> 2A=1/1.2-1/2015.2016

=> 2A < 1/2 => A < 1/4

nbvbvvvxcvcgf

A = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2015}\right)\)

A = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2014}{2015}\)

A = \(\frac{1}{2015}\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2015}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2014}{2015}=\frac{1\cdot2\cdot3\cdot...\cdot2014}{2\cdot3\cdot...\cdot2014\cdot2015}=\frac{1}{2015}\)

Câu 1: 

a: =(1+2-3-4)+(5+6-7-8)+...+(2013+2014-2015-2016)

=(-4)+(-4)+...+(-4)

=-4x504=-2016

b: \(B=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{195}{196}=\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot13\cdot15}{2\cdot3\cdot...\cdot14\cdot2\cdot3\cdot...\cdot14}=\dfrac{15}{14\cdot2}=\dfrac{15}{28}\)