1. -54+75-|-79-42|
2.2028- {[ 39 - 23.3-21]2 } :3+20170 }
3.47.134-47.35+47
4. -[-2017] +2789] + [1789 -2017 ]
5.369- [| -206| - 15] - [ -206 + |-369|
6. 345 - 150 : [{33 - 24}2 ] - [-21] + 20160
7.-2 +6 - 12 + 16 - 22 + 36 - ... - 92 + 96
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Câu 1 Tính
a) \(-54\) + 75 - \(\frac{1}{2}\) - 79 - 42\(\frac{1}{2}\)
= 21 - \(-78,5\)- 42\(\frac{1}{2}\)
= \( 99,5\) - 42\(\frac{1}{2}\)
= 57
Câu 2 Tính nhanh
a) 47 . 134 - 47 . 35 + 47
= 47 . ( 134 - 35 ) + 47
= 47 . 99 + 47
= 4653 + 47
= 4700
Câu 3:
a: =>(|x|+3)*15=75
=>|x|+3=5
=>x=2 hoặc x=-2
b: =>86:[2(2x-1)^2-7]=2*9-16=2
=>2(2x-1)^2-7=43
=>2(2x-1)^2=50
=>(2x-1)^2=25
=>2x-1=5 hoặc 2x-1=-5
=>x=-2 hoặc x=3
A.369 - (|-206| - 15 ) - ( - 206 + | -369|)
369-(206 - 15 ) + 206 - 369
= 369 - 206 + 15 + 206 - 369
= ( 369 - 369 ) + ( -206 + 206 ) + 15
= 0 + 0 + 15
= 15
B.345-150:[(3³-24)²-(-21)]+2016⁰
= 345-150:[(27-24)²+21]+1
= 345-150:(3²+21)+1
= 345-150:(9+21)+1
= 345-150:30+1
= 345-5+1
= 341
a, 369 - (|-206| - 15 ) - ( - 206 + | -369|)
369-(206 - 15 ) + 206 - 369
= 369 - 206 + 15 + 206 - 369
= ( 369 - 369 ) + ( -206 + 206 ) + 15
= 0 + 0 + 15
= 15
b, 345 - 150 : [ ( \(3^3\)- 24 )\(^2\)- ( -21 ) ] + 2016\(^0\)
= 345 - 150 : [(27 - 24)\(^2\) + 21] + 1
= 345 - 150 : [3\(^2\)+ 21 ] + 1
= 345 - 150 : 30 + 1
= 345 - 5 + 1
= 341
c, - 2 + 6 - 12 + 16 -....-92 + 96
ta có : ( 6 + 16 + ... + 96 ) + ( -2 + -12 + ... + -92 )
mỗi bên đều có 10 số hạng
=> [ ( 6 + 96 ) . 10 : 2 ] + [ ( -2 + -92 ) . 10 : 2 ]
=> 510 + -470
= 40
\(a,369-\left(\left|-206\right|-15\right)-\left(-206+\left|-369\right|\right)\)
\(=369-\left(206-15\right)+206-369\)
\(=369-206+15+206-369\)
\(=\left(369-369\right)+\left(-206+206\right)+15\)
\(=0+0+15=15\)
\(b,345-150:\left[\left(3^3-24\right)^2-\left(-21\right)\right]+2016^0\)
\(=345-150:\left[\left(27-24\right)^2+21\right]+1\)
\(=345-150:\left[3^2+21\right]+1\)
\(=345-150:30+1\)
\(=345-5+1=341\)
\(c,-2+16-12+...+96\)
Ta có: \(\left[-2+...+\left(-92\right)\right]+\left[6+16+...+96\right]\)
Hai bên có \(10\) số hạng.
\(\Rightarrow\left\{\left[-2+\left(-12\right)\right].10:2\right\}+\left\{\left[6+16\right].10:2\right\}\)
\(\Rightarrow510+\left(-470\right)=40\)
Bài 2:
a: =>20-16-x+6=90
=>10-x=90
=>x=-80
b: =>|x+8|=24-3*(32-25)=24-3*7=3
=>x+8=3 hoặc x+8=-3
=>x=-11 hoặc x=-5
c: \(\Leftrightarrow30+2^x-6=1000:25=40\)
=>2^x=16
=>x=4
c: =>x+2+9 chia hết cho x+2
mà x>=0
nên \(x+2\in\left\{3;9\right\}\)
hay \(x\in\left\{1;7\right\}\)
Câu 3 :
\(a,\left(\left|x\right|+3\right).15-5=70\)
=>\(\left(\left|x\right|+3\right).15=70+5=75\)
=>\(\left|x\right|+3=75:15=5\)
=>\(\left|x\right|=5-3=2\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy x ∈ {2;-2}
\(b,86:\left[2.\left(2x-1\right)^2-7\right]+42=2.3^2\)
=>86:[\(2.\left(2x-1\right)^2-7\)]\(=18-42\)
=>\(2.\left(2x-1\right)^2-7=-\frac{43}{12}\)
=>\(2.\left(2x-1\right)^2=-\frac{43}{12}+7=\frac{41}{12}\)
=>\(\left(2x-1\right)^2=\frac{41}{12}:2=\frac{41}{24}\)
=>\(2x-1=\sqrt{\frac{41}{24}}=\frac{\sqrt{246}}{12}\)
=>\(2x=\frac{\sqrt{246}}{12}+1=\frac{12+\sqrt{246}}{12}\)
=>\(x=\frac{12+\sqrt{246}}{12}:2=\frac{12+\sqrt{246}}{12.2}=\frac{12+\sqrt{246}}{24}\)
Vậy \(x\in\left\{\frac{12+\sqrt{246}}{24}\right\}\)
Câu 1:
\(a,54+75-\left|-79-42\right|\)
=\(129+121\)
=\(250\)
\(b,2048-\left\{\left[39-\left(2^3.3-21\right)^2\right]:3+2017^0\right\}\)
=\(2048-\left\{\left[39-\left(8.3-21\right)^2\right]:3+1\right\}\)
=\(2048-\left\{\left[39-9\right]:3+1\right\}\)
=\(2048-\left\{30:3+1\right\}\)
=\(2048-11\)
=\(2037\)
Câu 2:
\(a,47.134-47.35+47\)
=\(47.\left(134-35+1\right)\)
=\(47.100=4700\)
\(b,-\left(-2017+2789\right)+\left(1789-2017\right)\)
=\(2017+\left(-2789\right)+1789+\left(-2017\right)\)
=\(\left(2017-2017\right)+\left(-2789+1789\right)\)
=\(0+-1000=-1000\)
-54+75-|-79-42|
=-54+75-\(|-121|\)
=-54+75-121
=21-121
=-100
4/
-[-2017] +2789] + [1789 -2017 ]
=(-2017)+2789+1789-2017
=\([(-2017)+2017]\)+(2789-1789)
=0+1000
=1000