a, 16.(27+15)+8.(53+25):2

= 4.[(27+15).4] + 4.(53+25)

= 4.168 + 4.312

= 4.(168+312)

= 4.480=1920.

...

Câu b tình bt thui!

b, 53.(51+4)+53.53.(49+96)+53

=53.55+53.53.145+53

=2915+407305+53

=410220+53=410273

d,125.98.2.8.25

= 25.2.2.98.2.2.2.2.25.

= [(25.2.2.2)+(25.2.2.2)].98.

= (200 + 200).98.

= 400 . 98 = 39200.

1/53+-1/106+-1/159=|x|/318

6/318+-3/318+-2/318=|x|/318

1/318=|x|/318

=>|x|=1

x=1 hoặc x=-1

=> \(\frac{1}{53}\)+ \(\frac{-1}{106}\)+\(\frac{-1}{159}\)= \(\frac{\left|x\right|}{318}\)

=> \(\frac{1}{318}\)= \(\frac{\left|x\right|}{318}\)

=> x thuộc {1; -1}

\(\left(1-\frac{52}{53}\right)+\left(\frac{105}{106}-1\right)+\left(\frac{158}{159}-1\right)=\frac{\left|x\right|}{318}\)

\(\Rightarrow\frac{1}{53}+\frac{-1}{106}+\frac{-1}{159}=\frac{\left|x\right|}{318}\)

\(\Rightarrow\frac{1}{318}=\frac{\left|x\right|}{318}\)

\(\Rightarrow\left|x\right|=1\)

\(\Rightarrow x=\pm1\)

Vậy..............................

Lời giải:

$\frac{x-1001}{1002}+\frac{x-1950}{53}=\frac{x+158}{2161}+\frac{x+193}{2196}$

$\Leftrightarrow \frac{x-1001}{1002}-1+\frac{x-1950}{53}-1=\frac{x+158}{2161}-1+\frac{x+193}{2196}-1$

$\Leftrightarrow \frac{x-2003}{1002}+\frac{x-2003}{53}=\frac{x-2003}{2161}+\frac{x-2003}{2196}$

$\Leftrightarrow (x-2003)\left(\frac{1}{1002}+\frac{1}{53}-\frac{1}{2161}-\frac{1}{2196}\right)=0$

Dễ thấy $\left(\frac{1}{1002}+\frac{1}{53}-\frac{1}{2161}-\frac{1}{2196}\right)\neq 0$ nên $x-2003=0\Rightarrow x=2003$

bằng 10 nha mg

=1992 - 53 + 158 - 247 - 1592

=( 1992 - 1592 ) - [ 53  + 247 ] + 158

= 400 - 300 + 158

=100 + 158 = 258

1992 + ( - 53 ) + 158 + ( - 247 ) + ( - 1592 )

= 1992 + ( - 1592 ) + ( - 53 ) + ( - 247 ) + 158

= 400 + ( - 300 ) + 158

= 100 + 158

= 258

1992 + ( -53 ) + 158 + ( -247 ) + ( -1592 )

= 1992 - 53 + 158 - 247 - 1592

= ( 1992 - 1592 ) - ( 53 + 247 ) + 158

= 400 - 300 + 158

= 258

27 x (-53 ) + ( -27) x 47

= 27 x ( -53 + -47 )

= 27 x ( - 100 ) = -2700

a) -287 + 499 + (-499) + 285

=-287+285

=-2

b: Ta có: \(1992+\left(-53\right)+158+\left(-247\right)+\left(-1592\right)\)

\(=\left(1992-1592\right)+\left(-53-247\right)+158\)

\(=400-300+158=258\)

`[158-x]/31+[185-x]/29+[208-x]/27+[227-x]/25=10`

`<=>[158-x]/31-1+[185-x]/29-2+[208-x]/27-3+[227-x]/25-4=0`

`<=>[127-x]/21+[127-x]/29+[127-x]/27+[127-x]/25=0`

`<=>(127-x)(1/21+1/29+1/27+1/25)=0`

`<=>127-x=0`

`<=>x=127`