D

\(\left\{33-\left[16+\left(51:3\right)\right]\right\}+100\\ =\left\{33-\left[16+17\right]\right\}+100\\ =\left[33-33\right]+100\\ =0+100=100\)

\(51:3=17\) nha không phải \(51:2=17\)

\(\left\{33-\left[16+\left(51:3\right)\right]\right\}+100\\ =\left[33-\left(16+17\right)\right]+100\\ =\left(33-33\right)+100=0+100=100\)

Khi tới VN thì lúc đó giờ bên Mỹ là:

(19+12) - 24= 7 (giờ) (1/1/2020)

Vậy giờ bên Mỹ sơm hơn giờ ở Việt Nam là: 13 - 7 = 6 (tiếng)

=> 1h30 30/1/2020 ở HN thì Paris lúc đó đang là 19h30 29/1/2020

(Tuy nhiên Mỹ thường không sớm hơn VN 6 tiếng mà sớm hơn nhiều tiếng luôn ớ)

\(\dfrac{-12}{310}\text{ }và\text{ }\dfrac{8}{-36}\\ Có:310=2\cdot5\cdot31\\ 36=2^2\cdot3^2\\ \Rightarrow BCNN_{\left(310;36\right)}=2^2\cdot3^2\cdot5\cdot31=5580\\ \Rightarrow\dfrac{-12}{310}=\dfrac{-216}{5580}\\ \dfrac{8}{-36}=\dfrac{1240}{-5580}\\ mà\text{ }\dfrac{-216}{5580}>\dfrac{1240}{-5580}\\ Vậy\text{ }\dfrac{-12}{310}>\dfrac{8}{-36}\)

\(x^2\cdot y+x\cdot y-x=4\)

\(x\cdot y\cdot\left(x+1\right)-x=4\)

\(x\cdot y\cdot\left(x+1\right)-x-1=4-1\)

\(xy\cdot\left(x+1\right)-\left(x+1\right)=3\)

\(\left(x+1\right)\left(xy-1\right)=3\)

⇒ \((x + 1) ; (xy - 1)\) là ước của 3

⇒ \(\text{(x + 1) ; (xy - 1)}\in\left\{\pm1;\pm3\right\}\)

Ta có:

TH1:

\(\left[{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1-1\\xy=3+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\xy=4\end{matrix}\right.\Rightarrow y=\dfrac{xy}{x}=\dfrac{4}{0}\) (vô nghiệm)

TH2:

\(\left[{}\begin{matrix}x+1=-1\\xy-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1-1\\xy=-3+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\xy=-2\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{-2}{-2}=1\) (chọn)

TH3:

\(\left[{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-1\\xy=1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\xy=2\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{2}{2}=1\) (chọn)

TH4:

\(\left[{}\begin{matrix}x+1=-3\\xy-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3-1\\xy=-1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\xy=0\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{0}{-4}=0\)(vô nghiệm)

Vậy (x,y)ϵ{(-2 ; 1);(2 ; 1)}

tìm x, y thỏa mãn
x + 3 = y(x - 2)
=> x - 3 = y(x - 2) = 0
=> x - 3 = 0 hoặc y(x-2)=0
=> x = 3 và y = 0
Vậy x = 3 và y = 0

Nhớ tick

ủa nhưng mà nó là x-3 mà có phải là x+3 mà

e. = - 1080 + 120 

= -960 

f. (-41) x 59 + ( -41).2 + 59 x 41 - 59 x 2

= 2 x ( -41 - 59 ) 

= 2 x ( -100)

 = -200

g, ( 135 - 35 ) . ( -37 ) + 37 . ( -42 - 58 )

= 100 : ( -37 ) + 37 . -100

= 100 : ( -37 ) + ( 37 . -10 ) . -100

= 100 : ( -37 ) + ( -37 ) . -100

= ( -37 ) . ( 100 + -100 )

= ( -37 ) . 0

= 0

a)

\(\left(-11\right).\left(-28\right)+\left(-9\right).13\)

\(=308+\left(-9\right).13\)

\(=308+\left(-117\right)\)

\(=191\)

b)

\(\left(-69\right).\left(-31\right)-\left(-15\right).12\)

\(=\left(-69\right).\left(-31\right)+15.12\)

\(=2139+15.12\)

\(=2139+180\)

\(=2319\)

c)

\(\left[16-\left(-5\right)\right].\left(-7\right)\)

\(=\left[16+5\right].\left(-7\right)\)

\(=21.\left(-7\right)\)

\(=-147\)

d)

\(\left[\left(-4\right).\left(-9\right)-6\right].\left[\left(-12\right)-\left(-7\right)\right]\)

\(=\left[\left(-4\right).\left(-9\right)-6\right].\left[\left(-12\right)+7\right]\)

\(=\left[36-6\right].\left[\left(-12\right)+7\right]\)

\(=30.\left[\left(-12\right)+7\right]\)

\(=30.\left(-5\right)\)

\(=-150\)