Tìm x
x+(x+1)+(x+2)+...+2009+2010=2010
Tìm x thỏa mãn: x + (x + 1) + (x + 2) + … + 2009 + 2010 = 2010 A.-2010 B.-2008 C.0 D.-2009
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 : (2009-x)2+(2009-x)(x-2010)+(x-2010)2/(2009-x)2-(2009-x)(x-2010)+(x-2010)2=19/49
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 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
(2009-x)2+(2009-x)(2010-x)+(2010-x)2/(2009-x)2+(2009-x)(2010-x)+(2010-x)2 = 19/49
tìm x
Lời giải của mình ở đây nhé!
http://olm.vn/hoi-dap/question/424173.html
tìm x :
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{49}{19}\)
tìm x biết \(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
đặt 2009-x=a,x-2010=b
suy ra a^2+ab+b^2/a^2-ab+b^2=19/49
suy ra 49(a^2+ab+b^2)=19(a^2-ab+b^2)
49a^2+49ab+49b^2=19a^2-19ab+19b^2
30a^2+68ab+30b^2=0
30a^2+50ab+18ab+30b^2=0
10a(3a+5b)+6b(3a+5b)=0
(3a+5b)(10a+6b)=0
suy ra 3a+5b=0 hoặc 10a+6b=0
thế vào lại rồi tìm x