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a: =369-206+15+206-369=15

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

bỏ bớt ra ik, nhìn v nhìu quá

b: =100-60+17

=40+17

=57

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:

\(a,47.134-47.35+47\)

=\(47.\left(134-35+1\right)\)

=\(47.100=4700\)

\(b,-\left(-2017+2789\right)+\left(1789-2017\right)\)

=\(2017+\left(-2789\right)+1789+\left(-2017\right)\)

=\(\left(2017-2017\right)+\left(-2789+1789\right)\)

=\(0+-1000=-1000\)

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