a) \(A=2+2^2+2^3+...+2^{2017}\)

\(2A=2^2+2^3+2^4+...+2^{2018}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)

\(A=2^{2018}-2\)

b) \(C=1+3^2+3^4+...+3^{2018}\)

\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)

\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)

\(8C=3^{2020}-1\)

\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)

\(Toru\)

\(A=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right):2\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right):2\)

\(=\left(1-\frac{1}{2017}\right):2\)\(< \)\(\frac{1}{2}\)   (Do 1 - 1/2017 < 1)

\(A-B=35^2+33^2+31^2+....+3^2+1^2-\left(34^2+32^2+30^2+....+4^2+2^2\right)\\ =\left(35^2-34^2\right)+\left(33^2-32^2\right)+\left(31^2-30^2\right)+...+\left(3^2-2^2\right)+1^2\\ =\left(35-34\right)\left(35+34\right)+\left(33-32\right)\left(33+32\right)+\left(31-30\right)\left(31+30\right)+....+\left(3-2\right)\left(3+2\right)+1\\ =1.\left(35+34\right)+1.\left(33+32\right)+1.\left(31+30\right)+....+1.\left(3+2\right)+1\\ =1+2+3+....+30+31+32+33+34+35\\ =\dfrac{\left(1+35\right).35}{2}=630\)

\(=630\)

47;7;64)4)

B = 1 + 32 + 34 + … + 32018

32.B = 32.( 1 + 32 + 34 + … + 32018)

9B = 32 + 34 + 36 + … + 32020

9B – B = (32 + 34 + 36 + … + 32020) – (1 + 32 + 34 + … + 32018)

8B = 32020 – 1

B = (32020 – 1) : 8.

Vậy B = (32020 – 1) : 8.

tick cho mink nhé (●'◡'●)

B = (3²⁰²⁰ - 1) : 8

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

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

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

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

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

\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)

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

Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)

a/

Để biểu thức được xác định

\(=>\left\{{}\begin{matrix}2x-2\ne0\\2x+2\ne0\\x+1\ne0\end{matrix}\right.\)

\(\odot2x-2\ne0\)

\(2x\ne2\)

\(x\ne1\)

\(\odot2x+2\ne0\)

\(2x\ne-2\)

\(x\ne-1\)

\(\odot x+1\ne0\)

\(x\ne-1\)

Vậy điều kiện xác định của bt là: \(x\ne-1;x\ne\pm2\)

a: x là đơn thức một biến

b: A(x)=-x^2+2/3x-1

Đặt A(x)=0

=>-x^2+2/3x-1=0

=>x^2-2/3x+1=0

=>x^2-2/3x+1/9+8/9=0

=>(x-1/3)^2+8/9=0(vô lý)

c: B(-3)=(-3)^2+4*(-3)-5

=9-5-12

=4-12=-8

Ý bạn đề bài là: \(A=3+3^2+3^3+...+3^{14}\) phải không? Nếu đúng thì mình sẽ ghi lời giải ở dưới, sai đề thì bạn inbox riêng với mình nhé! Mình sẽ giải giúp bạn!

\(A=3+3^2+3^3+...+3^{14}\)

\(3A=3^2+3^3+...+3^{15}\)

\(3A-A=\left(3^2+3^3+...+3^{15}\right)-\left(3+3^2+...+3^{14}\right)\)

\(2A=3^{15}-3\)

\(A=\left(3^{15}-3\right):2\)

\(A=7174452\)

a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5