73.52+52.28-52
5x+53=1010
2727:(158-x)=27
tính
a, 16.(27+15)+8.(53+25):2
b, 53.(51+4)+53.53.(49+96)+53
c,158+445+555
d,125.98.2.8.25
e,7+10+13+16+19+22+25
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.
tính
a, 16.(27+75) +8 . (53+25):2
b, 53x ( 51+4) +53x (49+96) + 53
c,158+ 445 + 342+555
d, 125x98x2x8x25
e,7+10+13+16+19+22+25
dau cham la dau phay
Tìm x biết: (1-52/53) + (105/106 - 1) + ( 158/159 - 1) = |x|/318
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
Tìm x biết:
a, (1- 52/53)+(105/106-1)+(158/159-1)=|x|/318
=> \(\frac{1}{53}\)+ \(\frac{-1}{106}\)+\(\frac{-1}{159}\)= \(\frac{\left|x\right|}{318}\)
=> \(\frac{1}{318}\)= \(\frac{\left|x\right|}{318}\)
=> 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..............................
GIẢI PHƯƠNG TRÌNH
\(\frac{x-1001}{1002}+\frac{x-1950}{53}=\frac{x+158}{2161}+\frac{x+193}{2196}\)
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 $x-2003=0\Rightarrow x=2003$
giải pt sau :158-x/31 + 185-x/29 +208-x/27 + 277-x/25
1992+(-53)+158+(-247)+(-1592)
=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 ) + ( -27) x 47
= 27 x ( -53 + -47 )
= 27 x ( - 100 ) = -2700
1, Tính:
a) -287 + 499 + (-499) + 285
b) 1992 + (-53) + 158 + (-247) + (-1592)
c) 15 x (2 x 3\(^2\) - 4\(^2\)) + (7 x 2\(^0\)- 1) x (5\(^5\) : 5\(^4\) + 10\(^1\))
b: Ta có: \(1992+\left(-53\right)+158+\left(-247\right)+\left(-1592\right)\)
\(=\left(1992-1592\right)+\left(-53-247\right)+158\)
\(=400-300+158=258\)
Giải phương trình:
\(\dfrac{158-x}{31}+\dfrac{185-x}{29}+\dfrac{208-x}{27}+\dfrac{227-x}{25}=10\)
`[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`