a) 2005 - x + 2006 = 2007

2005 - x=2007-2006

2005 - x=1

x=2005-1

x=2004

b) 2003 < x + 200 < 2005

=>x+200=2004

x=2004-200

x=1804

c) x : 5 = 1402 ( dư 3 )

x=1402 x 5+3

x=7010 +3

x=7013

d) 409 : x = 5 ( dư 4 )

x=(409-4):5

x=81

e) 30 : 6 < x : 5 < 35 : 5

5 < x : 5 < 7

=>x:5=6

x=6 x 5

x=30

a)2005-x+2006=2007

  2005-x=2007-2006

2005-x=1

x=2005-1=2004

b)2003<x+200<2005

  2003<2004<2005

Vậy:x+200=2004

        x=2004-200=1804

c)x/5=1402(dư 3)

   x-3=5x1402

   x-3=7010

   x=7010+3=7013

    d)409/x=5 dư 4

      (409-4)/x=5

       405/x=5

      x=405/5=81

e)30/6<x/5<35/5

    5<x/5<7

     5<6<7

      x/5=6

     x=6x5=30

              a)2004;b)1804;c)7013;d)81;e)30

\(F=\left(x+1\right)^2+\left(2x-1\right)^2=x^2+2x+1+4x^2-4x+1=5x^2-2x+2=\left(x\sqrt{5}\right)^2-2x\sqrt{5}.\dfrac{1}{\sqrt{5}}+\dfrac{1}{5}+\dfrac{9}{5}=\left(x\sqrt{5}+\dfrac{1}{\sqrt{5}}\right)^2+\dfrac{9}{5}\ge0\)- minF=\(\dfrac{9}{5}\)⇔\(x\sqrt{5}+\dfrac{1}{\sqrt{5}}=0\)⇔x=\(\dfrac{-1}{5}\)

\(E=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\text{≥}-36\)  ∀x (vì \(\left(x^2+5x\right)^2\text{≥}0\))

MinE=-36 ⇔ \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=-3+3=0\)

\(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

=> x + 2010 = 0

=> x = -2010 

a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\) 

   8 .x + 1 . x = 990

x . [ 8 +1 ] = 990

x . 9 = 990

x = 990 : 9

x = 110

các bạn giúp mình với mình đang vội.

a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)

=>\(8\cdot x+1\cdot x=3305+1\)

=>\(9x=3306\)

=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)

b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)

=>\(2^x\left(1+2+4+8\right)=480\)

=>\(2^x\cdot15=480\)

=>\(2^x=32\)

=>\(2^x=2^5\)

=>x+5

a, \(\frac{4}{7}x\frac{5}{7}+\frac{3}{7}x\frac{5}{6}\)

= ( \(\frac{4}{7}+\frac{3}{7}\)) x \(\frac{5}{7}x\frac{5}{6}\)

= 1 x \(\frac{5}{7}x\frac{5}{6}\)

= \(\frac{5}{7}x\frac{5}{6}\)

= \(\frac{25}{42}\)

Tương tự mấy câu sau cũng làm như thế này