1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2009)(x+2010)
\(\dfrac{x+1}{2010}+\dfrac{x+2}{2009}+\dfrac{x-3}{2008}+...+\dfrac{x-2009}{2}+\dfrac{x-2010}{1}=-2010\)
\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)
=>x+2011=0
hay x=-2011
Tính hợp lý ( nếu được )
a) 2/9 x 11/5 - 1/3 x 7/15 b) 3/7 x 9/16 - 1/14 x 1/8
c) -1/2010 - 1/2010 x 2009 - 1/2009 x 2008 - .... - 1/3 x 2 - 1/2 x1
Tính nhanh:
A= 1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2009)(x+2010)
A= \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....\frac{1}{\left(x+2009\right)\left(x+2010\right)}\)
\(A=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x-1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}...\frac{1}{x+2009}-\frac{1}{x+2010}\)
\(A=\frac{1}{x}-\frac{1}{x+2010}\)
\(A=\frac{1}{x}+\frac{-1}{x+2010}\)
\(A=\frac{1\left(x+2010\right)}{x\left(x+2010\right)}+\frac{-1\cdot x}{x\left(x+2010\right)}\)
\(A=\frac{x+2010-x}{x\left(x+2010\right)}=\frac{2010}{x\left(x+2010\right)}\)
Giải PT:
a, \(\dfrac{x^2+x+1}{x^2+x+2}+\dfrac{x^2+x+2}{x^2+x+3}=\dfrac{7}{6}\)
b, \(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)
c, \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
Help me!!! Mk cần gấp!!!
đkxđ với mọi x
đặt a=x2+x+1
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)
=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)
<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14
<=> 12a2 -a2+24a-21a+6-14=0
<=> 11a2+3a-8=0
<=> 11a2 +11a-8a-8=0
<=> (11a2 +11a)-(8a+8)=0
<=> 11a(a+1)-8(a+1)=0
<=> (a+1)(11a-8)=0
=> a=-1 và a=\(\dfrac{8}{11}\)
thay a=x2+x+1 ta đc
x2+x+1=-1
<=> x2+x+2 =0 (vô nghiệm)
và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )
vậy pt trên vô nghiệm
c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0
( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)
\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)
\(< =>16=\left(x+4\right)^2\)
<=> x2 + 8x = 0
<=> x( x + 8) = 0
<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )
Vậy,....
Giúp mk câu a, c thui nha!! Câu b mk làm đc rùi!!!
Nhã Doanh, ngonhuminh, nguyen thi vang, @hattori heiji, @Phùng Khánh Linh, ...
Giải phương trình:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{^{x^2}}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (1 + 1/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (4/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x 0
=0
kết ban đi
đáp số: 0
dễ như ăn cháo
thảo mai ăn hủ tiếu mì nha
cho 2 đa thức P(x)=1+x+x^x+x^3+....+x^2009+x^2010 và Q(x)=1-x+x^2-x^3+x^4+....-x^2009+x^2010 giá trị của biểu thức P(1/2)+Q(1/2) có dạng biểu diễn hữu tỉ là a/b a,b thuộc N a,b là 2 số nguyên tố cùng nhau chứng minh a chia hết cho 5
giải phương trình: \(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2009}{2}+\frac{x+2010}{1}\)\(=\left(-2010\right)\)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
cách bạn giải bài này là gì vậy ?
Bài 1: Tìm x biết:
1/ x + (x + 1) + (x + 2) + (x + 3) +.....+ (x + 2006) + 2007 = 2007
2/ x + (x + 1) + (x + 2) +...+ 199 + 200 + 201 = 401
3/ x + (x + 1) + (x + 2) +....+ 2009 + 2010 = 2010
Bn nào nhanh mik tick nha!!!
1/x+x+1+x+2+x+3+...+x+2006+2007=2007
------------------------------------------=2007-2007
------------------------------------------=0
x+x+x+...+x+1+2+3+...+2006=0
2007.x+(1+2+...+2006)=0
2007.x+(2006+1).[(2006-1)+1]:2=0
2007.x+2013021=0
2007.x=0-2013021
x=-2013021:2007
x=-1003
2/x+x+1+x+2+...+x+198=401-201-200-199
199.x+(1+2+...+198)=-199
199.x+(1+198).[(198-1)+1]:2=-199
199.x+19701=-199
199.x=-199-19701
x=-19900:199
x=-100
3/x+x+1+x+2+...+x+2008=2010-2010-2009
2009.x+(2008+1).[(2008-1)+1]:2=-2009
2009.x+2017036=-2009
2009.x=-2009-2017036
x=-2019045:2009
x=-1005