73.52+52.28-52
5x+53=1010
2727:(158-x)=27
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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
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 $x-2003=0\Rightarrow x=2003$
=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
27 x (-53 ) + ( -27) x 47
= 27 x ( -53 + -47 )
= 27 x ( - 100 ) = -2700
b: Ta có: \(1992+\left(-53\right)+158+\left(-247\right)+\left(-1592\right)\)
\(=\left(1992-1592\right)+\left(-53-247\right)+158\)
\(=400-300+158=258\)
73.52+52.28-52
=52.(73+28-1)
=52.100
=25.100
=2500
5x+53=1010
5x+125=1010
5x=1010-125
5x=885
x=885:5
x=177
2727:(158-x)=27
158-x=2727:27
158-x=101
x=158-101
x=57
5x + 125 = 1010
5x = 1010 - 125
5x = 885
x = 885 : 5
x = 177