1992+(-53)+153+(-247)+(-1592)

=[1992+(-1592)]+[(-53)+153]+(-247)

=(-400)+100+(-247)

=(-300)+(-247)

=-547

`(x-2003)/16 +(x-1997)/11 +(x-1992)/9 +(x-1991)/7=10`

`<=>((x-2003)/16-1)+((x-1997)/11-2)+((x-1992)/9-3)+((x-1991)/7-4)=0`

`<=>(x-2019)/16+ (x-2019)/11 +(x-2019)/9+(x-2019)/7 =0`

`<=> (x-2019)(1/16+1/11+1/9+1/7)=0`

<=> x-2019=0`

`<=> x=2019`

a) 1012 = (100 + 1)2 = 1002 + 2.100 + 1 = 10000 + 200 + 1 = 10201

b) 1992 = (200 – 1)2 = 2002 – 2.200 + 1 = 40000 – 400 + 1 = 39601

c) 47.53 = (50 – 3)(50 + 3) = 502 – 32 = 2500 – 9 = 2491.

a) 39601.              b) 9801.             c) 249001.            d) 89999.

Ta có: \(\dfrac{1992x}{xy+1992x+1992}\)=

\(\dfrac{xyz.x}{xy+xyz.x+xyz}\) = \(\dfrac{xyz.x.z}{xy.z+xyz.x.z+xyz.z}\) = \(\dfrac{xz}{1+xz+z}\)

Ta có: \(\dfrac{y}{zy+y+1992}\)=\(\dfrac{y}{zy+y+xyz}\)=\(\dfrac{1}{z+1+xz}\)

=> \(\dfrac{1992x}{xy+1992x+1992}\)+\(\dfrac{y}{zy+y+1992}\)+\(\dfrac{z}{z+zx+1}\) = \(\dfrac{xz}{1+zx+z}\) +\(\dfrac{1}{z+zx+1}\) \(+\dfrac{z}{z+zx+1}\) =\(\dfrac{z+zx+1}{z+xz+1}\)

=1

Câu 1b :

\(\frac{x-2003}{16}+\frac{x-1997}{11}+\frac{x-1992}{9}+\frac{x-1991}{7}=10\)

\(\Leftrightarrow\) \(\left(\frac{x-2003}{16}-1\right)+\left(\frac{x-1997}{11}-2\right)+\left(\frac{x-1992}{9}-3\right)+\left(\frac{x-1991}{7}-4\right)=0\)

\(\Leftrightarrow\) \(\frac{x-2019}{16}+\frac{x-2019}{11}+\frac{x-2019}{9}+\frac{x-2019}{7}=0\)

\(\Leftrightarrow\) \(\left(x-2019\right)\left(\frac{1}{16}+\frac{1}{11}+\frac{1}{9}+\frac{1}{7}\right)=0\)

Mà \(\left(\frac{1}{16}+\frac{1}{11}+\frac{1}{9}+\frac{1}{7}\right)\ne0\)

\(\Rightarrow\) x - 2019 = 0 \(\Leftrightarrow\) x = 2019

\(\Rightarrow\) \(S=\left\{2019\right\}\)