Tìm x biết
a) |x-2|=|2x+3|
b) 2008 - |2x-2008|=2x
c) 2010 - |x-2010|=x
d) |x.(x-2)|=x(2-x)
1) TÌM X:
a) (2x-1)^2008=(2x-1)^2010
b)4^x -12*2^x+32=0 (đặt t=2^x>9)
2) Cho x^3-x=8 Tính A=x^6-2x^4-x+x^2+x^3
Tìm x biết:
a) x+2x+3x+4x+......+2011x=2012.2013
b)(x-1)/2011+(x-2)/2010-(x-3)/2009=(x-4)/2008
a) x+2x+3x+4x+...+2011x = 2012.2013
\(\Rightarrow\) x(1+2+3+4+...+2011) = 4050156
\(\Rightarrow\) x.2023066 = 4050156
\(\Rightarrow\) x = 4026/2011
Câu a ko nhất thiết phải tính ra số lớn như thế đâu
Bài 1: cho pt \(x^2-ax+a-1=0\) có 2 no x1, x2
Tính \(M=\dfrac{2x^2_1+x_1x_2+2x_1^2}{x^2_1x_2+x^2_2x_1}\)
Bài 2: cho a,b là no pt: \(30x^2-4x=2010\)
Tình \(N=\dfrac{30\left(a^{2010}+b^{2010}\right)-4\left(a^{2009}+b^{2009}\right)}{a^{2008}+b^{2008}}\)
Bài 2:
Vì a,b là nghiệm PT nên \(\left\{{}\begin{matrix}30a^2-4a=2010\\30b^2-4b=2010\end{matrix}\right.\)
\(\Rightarrow N=\dfrac{a^{2008}\left(30a^2-4a\right)+b^{2008}\left(30b^2-4b\right)}{a^{2008}+b^{2008}}\\ \Rightarrow N=\dfrac{a^{2008}\cdot2010+b^{2008}\cdot2010}{a^{2008}+b^{2008}}=2010\)
Bài 1:
Viét: \(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=a-1\end{matrix}\right.\)
\(M=\dfrac{2x_1^2+x_1x_2+2x_2^2}{x_1^2x_2+x_1x_2^2}=\dfrac{2\left(x_1+x_2\right)^2-3x_1x_2}{x_1x_2\left(x_1+x_2\right)}=\dfrac{2a^2-3a+3}{a^2-a}\)
giải các ptr sau
a)\(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)
b)\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)
a)3,6-0,5(2x-1)=3x-0,25(3-4x)
b)5x^2-4x-1=0
c)2-x/2008-1=1-x/2009-x/2010
a: =>3,6-x+0,5=3,5-0,75+x
=>4,1-x=x+2,75
=>-2x=-1,35
=>x=0,675
b: =>5x^2-5x+x-1=0
=>(x-1)(5x+1)=0
=>x=1 hoặc x=-1/5
c: \(\Leftrightarrow\left(\dfrac{2-x}{2008}+1\right)=\left(\dfrac{1-x}{2009}+1\right)+\left(1-\dfrac{x}{2010}\right)\)
=>\(2010-x=0\)
=>x=2010
Tìm x; y; z:
a) |x - 2| - |2x+3| - x = 0
b) |x - 7| + 2x+5=6
c)(3x-5)2006+(y2-1)2008+(x-z)2010 = 0
d) 2009- |x - 2009| = x
e)(2x-1)2008+(y- \(\frac{2}{5}\))2008+ |x+y - z| =0
Tìm x thỏa mãn: x + (x + 1) + (x + 2) + … + 2009 + 2010 = 2010 A.-2010 B.-2008 C.0 D.-2009
Tìm x,y thuộc Z:
a, (x-3)^2+(y+2)^2=0
b,2x+2^x+3=136
c,42-3./y-3/=4.(2042-x)^4
d,/x+5/+(3y-6)^2010=0
e,(2x-4)^2008+(y-4)^2008+/x+y+z/=0
g,(3x-6)^2006+(y^2-1)^2008+(x-z)^2100=0
h,8.2^3x.7^y=56^2x.5^x-1
i, x^3-y^3-z^3=3xyz và x^2=2.(y+z) (x,y,z thuộc N*)
\(\dfrac{x+1}{2010}+\dfrac{x+2}{2009}+\dfrac{x-3}{2008}+...+\dfrac{x-2009}{2}+\dfrac{x-2010}{1}=-2010\)
\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)
=>x+2011=0
hay x=-2011