`#3107.101107`

`-3^2 + {-54 \div [-2^8 + 7] * (-2)^2}`

`= -9 + [-54 \div (-256 + 7) * 4]`

`= -9 + [-54 \div (-249) * 4]`

`= -9 + (18/83 * 4)`

`= -9 + 72/83`

`= -675/83`

______

`31 * (-18) + 31 * (-81) - 31`

`= 31 * (-18 - 81 - 1)`

`= 31 * (-100)`

`= -3100`

___

`(-12) * 47 + (-12) * 52 + (-12)`

`= (-12) * (47 + 52 + 1)`

`= (-12) * 100`

`= -1200`

___

`13 * (23 + 22) - 3 * (17 + 28)`

`= 13 * 45 - 3 * 45`

`= 45 * (13 - 3)`

`= 45 * 10`

`= 450`

____

`-48 + 48 * (-78) + 48 * (-21)`

`= 48 * (-1 - 78 - 21)`

`= 48 * (-100)`

`= -4800`

Ta có: \(\left( C \right):{\left( {x + 6} \right)^2} + {\left( {y - 12} \right)^2} = 81 \Leftrightarrow {\left( {x - \left( { - 6} \right)} \right)^2} + {\left( {y - 12} \right)^2} = {9^2}\)

=> \(I\left( { - 6;12} \right)\) .

Chọn B

Chứng minh gì bạn?

\(\left\{{}\begin{matrix}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+3y+5x-5y=12\\-5x-5y+2x-2y=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}8x-2y=12\\-3x-7y=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}8x-2y=12\\3x+7y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-6y=36\\24x+56y=-96\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-62y=132\\8x-2y=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{66}{31}\\4x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{66}{31}\\4x=y+6=\dfrac{120}{31}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{30}{31}\\y=-\dfrac{66}{31}\end{matrix}\right.\)

a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)

\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)

\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)

\(Q=\left(x-y-2x-4y\right)^2\)

\(Q=\left(-x-5y\right)^2\)

b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)

\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)

\(A=\left[\left(xy+2\right)-2\right]^3\)

\(A=\left(xy+2-2\right)^3\)

\(A=\left(xy\right)^3\)

\(A=x^3y^3\)

c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)

\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)

\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)

\(=0\)

a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2

=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2

b: =(xy+2-2)^3=(xy)^3=x^3y^3

c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)

=24x+2x^3-2x^3-24x

=0

a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)

b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)

c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)

Ta có:

|12-x|=-|x-12|

Đ k: x-12>_0=>x>_12

=>2014.|x-12|+(x-12)^2=-2013.|x-12|

=>2014.(x-12)+(x-12)^2+2013.(x-12)=0

=>(x-12).(2014+x-12+2013)=0

=>(x-12).(x+2005)=0

=>x-12=0 và. x+2005=0

=>x=12 và x=-2005

a)

4 . 25 – 12 . 25 + 170 : 10 

= (4 . 25) – (12 . 25) + (170 : 10) 

= 100 - 300 + 17 

= -183

b)

(7 + 33 + 32) . 4 – 3 

= (7 + 27 + 9) .4 – 3

= 43 . 4 – 3

= (43 . 4) – 3

= 45

c)

12 : {400 : [500 – (125 + 25 . 7)}

= 12 : {400 : [500 – (125 + 175)}

= 12 : (400: 200)

= 12 : 2

= 6

d)

168 + {[2.(2⁴ + 3²) - 256⁰] : 72}.

= 168 + [2 . (16 + 9) – 1] : 49

= 168 + 49: 49

= 168 + 1 

= 167

a) 

4 . 25 – 12 . 25 + 170 : 10 

= (4 . 25) – (12 . 25) + (170 : 10) 

= 100 - 300 + 17 

= -183

b)

(7 + 33 + 32) . 4 – 3 

= (7 + 27 + 9) .4 – 3

= 43 . 4 – 3

= (43 . 4) – 3

= 45

c)

12 : {400 : [500 – (125 + 25 . 7)}

= 12 : {400 : [500 – (125 + 175)}

= 12 : (400: 200)

= 12 : 2

= 6

d)

168 + {[2.(2⁴ + 3²) - 256⁰] : 72}.

= 168 + [2 . (16 + 9) – 1] : 49

= 168 + 49: 49

= 168 + 1 

= 167

a)

4.25 -12.5 + 170 : 10=100−60+17=40+17=57.​

b)

 (3:3((7+33:32).4−3=(7+3).4−3=40−3=37.

c)

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