a)=(3/8+10/16)+(7/12+10/24)

=1+1=2

c)=(4/6+14/6)+(7/13+19/13)+(17/9+1/9)

=3+2+2=7

\(\frac{2006\times2004-9}{1995+2004\times2005}=\frac{\left(2005+1\right)\times2004-9}{1995+2004\times2005}\)

\(=\frac{2005\times2004+2004-9}{2004\times2005+1995}\)

\(=\frac{2005\times2004+1995}{1995+2004\times2005}\)

\(=1\)

2006x 2004 -9/1995 +2004x 2005

=(2005+1)x2004 -9/1995+2004x2005

=2005x2004+2004 x1-9/1995+2005x2004

=2005x2004+2004-9/1995+2004x2005

=2005x2004+1995/1995+2005x2004

=1

2006x 2004-9/ 1995+2004x 2005

=(2005+1)x 2004- 9/ 1995+2004x2005

= 2005x2004+2004-9(nhân phân phối)/ 1995+2004x2005

=2005x2004+1995/ 1995+2004x2005

=1(cả mẫu lẫn tử đều giống nhau nên thu gọn)

click cho tớ nhé

\(2004\)x\(2006-\frac{2003}{2005}\)x\(2005-2004\)

\(=2004\)x\(2006-\frac{2003}{2005}\)x\(\frac{2005}{1}-2004\)

\(=2004\)x\(2006-2003-2004\)

\(=2006-2003\)

\(=3\)

   2004 x 2006 - 2003 / 2005 x 2005 - 2004

= 2004 x ( 2005 + 1 ) - 2003 / ( 2004 + 1 ) x 2005 - 2004

= 2004 x 2005 + 2004 x 1 - 2003 / 2004 x 2005 + 1 x 2005 - 2004

= 2004 x 2005 + 2004 - 2003 / 2004 x 2005 + 2005 - 2004

= 2004 x 2005 + 1 / 2004 x 2005 + 1

= 1 / 1 = 1

Nếu có sai chỗ nào thì chỉ mk nha !

Hok tốt !

a)    \(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=0\)

\(\Leftrightarrow\)\(\frac{x+1}{4}-1-\frac{x+2}{5}+1+\frac{x+4}{7}-1-\frac{x+5}{8}+1+\frac{x+7}{10}-1-\frac{x+9}{12}+1=0\)

\(\Leftrightarrow\)\(\frac{x-3}{4}-\frac{3-x}{5}+\frac{x-3}{7}-\frac{3-x}{8}+\frac{x+3}{10}-\frac{3-x}{12}=0\)

\(\Leftrightarrow\)\(\frac{x-3}{4}+\frac{x-3}{5}+\frac{x-3}{7}+\frac{x-3}{8}+\frac{x-3}{10}+\frac{x-3}{12}=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right)=0\)

Vì   \(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\ne0\)

\(\Rightarrow\)\(x-3=0\)

\(\Leftrightarrow\)\(x=3\)

Vậy...

b)   \(\frac{x}{2004}+\frac{x+1}{2005}+\frac{x+2}{2006}+\frac{x+3}{2007}=4\)

\(\Leftrightarrow\)\(\frac{x}{2004}-1+\frac{x+1}{2005}-1+\frac{x+2}{2006}-1+\frac{x+3}{2007}-1=0\)

\(\Leftrightarrow\)\(\frac{x-2004}{2004}+\frac{x-2004}{2005}+\frac{x-2004}{2006}+\frac{x-2004}{2007}=0\)

\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)

Vì   \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\)

\(\Rightarrow\)\(x-2004=0\)

\(\Leftrightarrow\)\(x=2004\)

Vậy...

\(\frac{2006\times2004-9}{2004\times2005+1995}=\frac{2005\times2004+2004-9}{2004\times2005+1995}=\frac{2005\times2004+1995}{2004\times2005+1995}=1\)

ket qua la 1

a) \(C=x^3+3x^2+3x+10=\left(x+1\right)^3+9\)

Tại x = 99...9 (2004 chữ số 9) thì: x+1 = 100...0 (2004 chữ số 0) = 10²⁰⁰⁴

Khi đó, C = (10²⁰⁰⁴)³ + 9 = 10⁶⁰¹² + 9.

b) \(B=\left(5x-11\right)^2-\left(10x-22\right)\left(5x-9\right)+\left(5x-9\right)^2=\)

\(=\left(5x-11\right)^2-2\cdot\left(5x-11\right)\left(5x-9\right)+\left(5x-9\right)^2=\left(5x-11-\left(5x-9\right)\right)^2=\left(-2\right)^2=4\)

Hay B = 4 với mọi x .

Vậy tại x = 2005²⁰⁰⁶ thì B = 4.

a/ /1-2x/=125+75=200

=> 1-2x=200 hoac 1-2x=-200

=> x=-199/2 hoac x=201/2

b/ /x+4/=5-8+2004=2001

=> x+4=2001 hoac x+4=-2001

=> x=1992 hoac x=-2005

c/ /x-9/-1=(-45)-81+2006=1880

=> /x-9/ = 1880+1=1881

=> x-9=1881 hoac x-9=-1881

=> x=1990 hoac x=-1872