5x +x=39-3 mũ 11: 3 mũ 9
giúp mình với
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\(B=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(B=\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^2\cdot10\cdot\left(3^2\right)^4}\)
\(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^9\cdot10\cdot3^8}\)
\(B=\dfrac{2^{11}\cdot2^2\cdot3^6+2^{11}\cdot3^6\cdot3^3}{2^{11}\cdot2^3\cdot3^6\cdot3+\dfrac{2^{11}}{2^2}\cdot10\cdot3^6\cdot3^2}\)
\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3\right)+2^{11}\cdot\dfrac{1}{2^2}\cdot10\cdot3^6\cdot3^2}\)
\(B=\dfrac{\left(2^{11}\cdot3^6\right)\left(2^2+3^3\right)}{\left(2^{11}\cdot3^6\right)\left(2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2\right)}\)
\(B=\dfrac{2^2+3^3}{2^3\cdot3+\dfrac{1}{2^2}\cdot10\cdot3^2}\)
\(B=\dfrac{4+27}{8\cdot3+\dfrac{1}{4}\cdot10\cdot9}\)
\(B=\dfrac{31}{24+\dfrac{1}{4}\cdot90}\)
\(B=\dfrac{31}{24+\dfrac{45}{2}}\)
\(B=\dfrac{31}{\dfrac{48}{2}+\dfrac{45}{2}}\)
\(B=\dfrac{31}{\dfrac{93}{2}}\)
\(B=31\div\dfrac{93}{2}\)
\(B=31\times\dfrac{2}{93}\)
\(B=\dfrac{2}{3}\)
\(\left\{33-\left[16+\left(51:3\right)\right]\right\}+100\\ =\left\{33-\left[16+17\right]\right\}+100\\ =\left[33-33\right]+100\\ =0+100=100\)
\(51:3=17\) 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
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
5x + x = 39 - 311 : 39
6x = 39 - 32 = 39 - 9 = 30
x = 30 : 6 = 5