hnuji9on ui bm, 76tfv45tj,

x+3=0

=>x=-3

Vậy A=-3²⁰¹²+5²⁰¹¹

 ta có :

Đọc tiếp...

Có lẽ là đề sai, đề đúng phải là \(x=\frac{\sqrt{5}-1}{2}\)

Khi đó \(2x+1=\sqrt{5}\Rightarrow4x^2+4x+1=5\Leftrightarrow x^2+x-1=0\)

\(A=\frac{\left(x^2+x-1-2\right)^{2011}}{\left(x^3\left(x^2+x-1\right)-2\right)^{2011}}+\left(x^3\left(x^2+x-1\right)+1\right)^{2011}\)

\(A=\frac{\left(-2\right)^{2011}}{\left(-2\right)^{2011}}+1^{2011}=2\)

Làm theo cách này nhé : 

a = 2 / 60 x 63 + 2 / 63 x 66 + 2 / 66x 69 + ...+ 2 / 117 x 120 + 2 / 2011

= 2/3 x ( 3/60 x 63 +  3 / 63 x 66 + 3 / 66  x 69 + ...+ 3/117 x 120 ) + 2/2011

= 2/3 x ( 1/60 - 1/63 + 1/63 - 1/66 + 1/66 - 1/69 + ... + 1/117 - 1/120 ) + 2/2011

= 2/3 x  (  1/60 - 1/120 ) + 2/2011

= 2/3 x   1/120 + 2/2011

= 1/180 + 2/2011

b =  5/ 40 x 44 + 5 / 44 x 48 + ...+ 5/76 x 80 + 5/ 2011

= 5/4 x ( 4/40 x 44 + 4/44 x 48 + ...+ 4/76 x 80 ) + 5/2011

= 5/4  x ( 1/40 - 1/44 + 1/44 - 1/48 + ...+ 1/76 - 1/80 ) + 5/2011

= 5/4 x ( 1/40 - 1/80 ) + 5/2011

= 5/4 x        1/80 + 5/2011

= 1/64 + 5/2011

Do 1/64 > 1/80 ;   5/2011 > 2/2011

=> 1/64 + 5/2011 > 1/80 + 2/2011

=> b > a

K nha

Mình sửa lại chút nhé , lỗi đánh bàn phím thoy , :

Do 1/64 > 1/180 ;   5/2011 > 2/2011

=> 1/64 + 5/2011 > 1/180 + 2/2011

=> b > a 

b >a  bạn nhé

a ) \(2011^2.2011^x=2011^6\)

   \(2011^{x+2}=2011^6\)

=> x + 2 = 6 

=> x = 4 

b) \(5^x+1=126\)

  \(5^x=125\)

 \(5^x=5^3\)

=> x  = 3 

a) 100 - 99 + 98 -97 + 96 -95 +...+ 4-3 + 2

= (100 - 99) + (98 -97) + (96 - 95) +...+ (4-3) +2 (gồm 49 cặp và 1 số hạng)

= 1+1+1+....+1 +2

= 49 x 1 + 2 = 51

b) 100 - 5-...-5 - 5  (20 số 5)

= 100 - 20 x 5 = 0

c) 99 - 9 - 9 -... - 9 -9 (11 số 9)

=99 - 11 x 9 = 0

d) 2011 + 2011+2011+2011 - 2008 x 4

= 2011 x 4 - 2008 x 4

= 4 x (2011 - 2008)

= 4 x 3

=12

x²+y²+z²= xy+yz+zx.

=> 2x²+2y²+2z²-2xy-2yz-2zx=0

=> ( x-y)²+(y-z.)²+(z-x)² =0

=> x=y=z=0

Thay x=y=z vào x²⁰¹¹+y²⁰¹¹+z²⁰¹¹=3²⁰¹² ta được:

3.x²⁰¹¹=3.3²⁰¹¹

=> x²⁰¹¹=3²⁰¹¹

=> x=3 hoặc x = -3

Hay x=y=z=3 hoặc x=y=z=-3

1) có bn giải rồi ko giải nữa

2) \(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2011^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2012^4+\frac{1}{4}\right)}\)

Với mọi n thuộc N ta có :

\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)

\(=\left[n\left(n-1\right)+\frac{1}{2}\right]\left[n\left(n+1\right)+\frac{1}{2}\right]\)

Áp dụng ta được :

\(A=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(2011.2012+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right).......\left(2012.2013+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}}{2012.2013+\frac{1}{2}}=\frac{1}{8100313}\)

Ta có : \(\frac{x+2011}{1}+\frac{x+2008}{2}+\frac{x+2007}{3}+\frac{x+2011}{5}=-15\)

\(\Rightarrow\left(\frac{x+2011}{1}+5\right)+\left(\frac{x+2008}{2}+4\right)+\left(\frac{x+2007}{3}+3\right)+\left(\frac{x+2008}{4}+2\right)+\left(\frac{x+2011}{5}+1\right)\)

\(=0\)

=> \(\frac{x+2016}{1}+\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{4}+\frac{x+2016}{5}=0\)

=> \(\left(x+2016\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=0\)

Vì \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\ne0\)

=> x + 2016 = 0

=> x = -2016

Vậy x = -2016

x+(x+2008)1/2+(x+2007)1/3+(x+2008)1/4+(x+2011)1/5=-15-2011=-2026

<=> x+x/2+1004+x/3+669+x/4+502+x/5+2011/5=-2026

<=>x+x/2+x/3+x/4+x/5+2011/5=-2026-1004-669-502=-4201

<=>x(1+(1)/(2)+(1)/(3)+(1)/(4)+(1)/(5))=-4201-(2011)/(5)=-23016/5

<=>x=-23016/5:(1+1/2+1/3+1/4+1/5)=-2016

           Bài làm :

Ta có :

 \(\frac{x+2011}{1}+\frac{x+2008}{2}+\frac{x+2007}{3}+\frac{x+2011}{5}=-15\)

\(\Rightarrow\left(\frac{x+2011}{1}+5\right)+\left(\frac{x+2008}{2}+4\right)+\left(\frac{x+2007}{3}+3\right)+\left(\frac{x+2008}{4}+2\right)+\left(\frac{x+2011}{5}+1\right)\)

\(=0\)

 \(\Leftrightarrow\frac{x+2016}{1}+\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{4}+\frac{x+2016}{5}=0\)

 \(\Leftrightarrow\left(x+2016\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=0\)

 \(\text{Vì : }1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}>0\)

\(\Rightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

Vậy x=-2016

a: \(\Leftrightarrow\left(\left|x\right|\right)^2-5\left|x\right|-6=0\)

\(\Leftrightarrow\left(\left|x\right|-6\right)\left(\left|x\right|+1\right)=0\)

\(\Leftrightarrow\left|x\right|-6=0\)

=>x=6 hoặc x=-6

b: \(\dfrac{x}{x-2}+\dfrac{5}{\left|x+2\right|}=1\)

Trường hợp 1: x>-2 và x<>2

Pt sẽ là \(\dfrac{x}{x-2}+\dfrac{5}{x+2}=1\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=x\left(x+2\right)+5\left(x-2\right)\)

\(\Leftrightarrow x^2+2x+5x-10=x^2-4\)

=>7x=6

hay x=6/7(nhận)

TRường hợp 2: x<-2

Pt sẽ là \(\dfrac{x}{x-2}-\dfrac{5}{x+2}=1\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=x\left(x+2\right)-5\left(x-2\right)\)

\(\Leftrightarrow x^2+2x-5x+10=x^2-4\)

=>-3x=-14

hay x=14/3(loại)