tìm x biết : |x-2009|-|2009-x|=0
Ai giúp mình với,cô cho toàn bài khó.
B1:
a)Tìm x,y biết (x+y)^2=(x-1)(y+1)
b)Tìm x,y,z biết :9x^2+y^2+2z^2-18x+4z-6y +20=0
B2:
Cho x/a+y/b+z/c=1 và-a/x+b/y+c/z=0
C/m x^2/a^2 +y^2/b^2 +z^2/c^2=1
B3:
Tìm x
(2009-x)^2+(2009-x)(x-2010)+(x-2010)^2/(2009-x)^2-(2009-x)(x-2010)+(x-2010)^2=19/49
Tìm x:
x ( x - 2009 ) - 2010x + 2009 x 2010 = 0
\(x.\left(x-2009\right)-2010x+2009.2010=0\)
\(x.\left(x-2009\right)-2010\left(x-2009\right)=0\)
\(\left(x-2009\right)\left(x-2010\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2009=0\\x-2010=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=2010\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=2009\\x=2010\end{cases}}\)
a) 25 - y^2 = 8(x+2009)^2 \Leftrightarrow 8(x+2009)^2 + y^2 = 25
Do y^2 \geq 0 \Rightarrow (x+2009)^2 \leq 25/8
\Rightarrow x+2009 =0 hoặc 1
Nếu x+2009 = 1 \Rightarrow 25 - y^2 = 1\Rightarrow y^2 = 26 (không tìm được y)
Nếu x+2009 = \Rightarrow 25 - y^2 = 0\Rightarrow y^2 = 25, y=5
Vậy (x=0;y=5)
tìm x biết :
(2009 - x^2) + ( 2009 - x^2)( x - 2010) + ( x - 2010) / (2009 - x^2) - ( 2009 - x^2)( x - 2010) + ( x - 2010 ) = 19/49
tìm x biết 2009 - |x-2009| = x
2009-|(x-2009)|=x
=> |x-2009|=2009-x
TH1: x-2009 = 2009-x
=> x+x= 2009+2009
=> x=2009
TH2: -x+2009=2009-x
=> -x+x=2009-2009
=> 0=0
vậy x= 2009
Tìm x biết:
a) |x|+3√x^2+4+x^2010=6
b) 2010|7-x^2|^2009+2009(x+√7)^2010≤0
(Mình cần gấp lắm, giúp mình nha)
Tìm x biết: (2009-x)2 + (2009-x)(x-2010) + (x-2010)2/(2009-x)2 - (2009-x)(x-2010) + (x-2010)2 = 19/49
dat a =2009-x
b=x-2010
ta co : a^2+ab+b^2/a^2-ab+b^2 =19/49
<=>49a^2+49ab+49b^2=19a^2-19a+19b^2
<=>30a^2+68a+30b^2=0
<=>15a^2+34ab+15b^2=0
<=>15a^2+9ab+25ab+15b^2=0
<=>3a(5a+3b)+5b(5a+3b)=0
<=>(5a+3b)(3a+5b)=0
<=>5a+3b=0 hoac 3a+5b=0
vs 5a +3b=0 <=>5(2009-x)+3(x-2010)=0=>x=......
Tìm x biết: (2009-x)2 + (2009-x)(x-2010) + (x-2010)2/(2009-x)2 - (2009-x)(x-2010) + (x-2010)2 = 19/49
tìm x biết : (x-2009)(x-2009)(x-2007)...(x+2006)(x+2010)