nhanh hộ sunny vs các tềnh yew ới ời ơi !

\(P=x^3+2x^2+x+1\)

Khi \(x=3\Rightarrow P=3^3+2.3^2+3+1=49\)

Ta có:

\(8=2^3\)

\(12=2^2\cdot3\)

\(15=3\cdot5\)

\(\Rightarrow BCNN\left(8;12;15\right)=2^3\cdot3\cdot5=120\)

\(\Rightarrow BC\left(8;12;15\right)=\left\{0;120;240;360;480;...\right\}\)

Nhanh lên giúp mình nhé

Cảm ơn 

Các dữ liệu đê fbafi chưa đầy đủ em nhé, em xem lại đề bài xem đã đăng đúng và đầy đủ chưa?

          Cách 1:

     (a\(x^2\) + b\(x\) + c).(\(x+3\))

= a\(x^3\) + 3a\(x^2\) + b\(x^2\) + 3b\(x\) + c\(x\) + 3c

= a\(x^3\) + (3a\(x^2\) + b\(x^2\)) + (3b\(x\) + c\(x\)) + 3c

= a\(x^3\) + \(x^2\).(3a + b) + \(x\).(3b + c) + 3c

a\(x^3\) + (3a + b)\(x^2\) + (3b + c)\(x\) + 3c = \(x^3\) + 2\(x^2\) - 3\(x\)

⇔ \(\left\{{}\begin{matrix}a=1\\3a+b=2\\3b+c=-3\\3c=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}a=1\\3+b=2\\3b+c=-3\\c=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}a=1\\b=2-3\\3b=-3\\c=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}a=1\\b=-1\\b=-1\\c=0\end{matrix}\right.\)

Vậy (a; b; c) = (1; -1; 0)

         Cách hai ta có:

\(x^3\) + 2\(x^2\) - 3\(x\) = (\(x^3\) + 3\(x^2\)) - (\(x^2\) + 3\(x\))

\(x^3\) + 2\(x^2\)  - 3\(x\) = \(x^2\).(\(x+3\)) - \(x\).(\(x+3\))

\(x^3\) + 2\(x^2\) - 3\(x\) = (\(x+3\)).(\(x^2\) - \(x\))

⇒ (a\(x^2\) + b\(x\) + c).(\(x\) + 3) = (\(x+3\)).(\(x^2\) - \(x\))

⇔ a\(x^2\) + b\(x\) + c = \(x^2\) - \(x\)

⇒ \(\left\{{}\begin{matrix}a=1\\b=-1\\c=0\end{matrix}\right.\)

Vậy (a; b; c) = (1; -1; 0)

giờ đầu đi được: 91,8 x \(\dfrac{1}{3}\) = 30,6 (km)
Quãng đường còn lại: 91,8 - 30,6 = 61,2 (km)
giờ thứ hai đi được: 61,2 x \(\dfrac{2}{3}\) = 40,8 (km)
Quãng đường giờ thứ ba xe đi đươc: 91,8 - 30,6 - 40,8 = 20,4 (km)
Đs: giờ thứ ba: 20,4 km
 

`((-27)^10 . 16^25)/(6^30 . (-32)^15)`
`= (3^30 . 2^100)/(2^30 . 3^30 . -2^75)`
`= (2^100)/(2^30 . -2^75)`
`= (2^70)/(-2^75)`
`= -2^145`

A = \(\dfrac{\left(-27\right)^{10}.16^{25}}{6^{30}.\left(-32\right)^{15}}\)

A = \(\dfrac{\left(3^3\right)^{10}.\left(2^4\right)^{25}}{\left(2.3\right)^{30}.\left(-32\right)^{15}}\)

A = \(\dfrac{3^{30}.2^{100}}{2^{30}.3^{30}.\left(-2^5\right)^{15}}\)

A = \(\dfrac{3^{20}.2^{100}}{3^{30}[2^{30}.\left(-2\right)^{75}].}\)

A = \(-\dfrac{2^{100}}{\left[2^{30}.\left(2\right)^{75}\right]}\)

A = -2^{100 - 30 - 75}

A = - 2^70-75

A = -2^-5 

A = - \(\dfrac{1}{32}\)

250

\(\left(x+\dfrac{1}{3}\right)^3=\left(\dfrac{2}{3}\right)^6\\ \Rightarrow\left(x+\dfrac{1}{3}\right)^3=\left[\left(\dfrac{2}{3}\right)^2\right]^3\\ \Rightarrow\left(x+\dfrac{1}{3}\right)^3=\left(\dfrac{4}{9}\right)^3\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{4}{9}\\ \Rightarrow x=\dfrac{4}{9}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{4}{9}-\dfrac{3}{9}\\ \Rightarrow x=\dfrac{1}{9}\)

\(A=\left(9.5\right)^{40}-5^{40}=9^{40}.5^{40}-5^{40}\)

\(=5^{40}.\left(9^{40}-1\right)=5^{2.20}.\left(9^{40}-1\right)=\left(5^2\right)^{20}.\left(9^{40}-1\right)\)

\(=25^{20}.\left(9^{40}-1\right)⋮25^{20}\)

`#3107.101107`

\(\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{2}{3}\right)^6\\ \left(x+\dfrac{1}{3}\right)^2=\left[\pm\left(\dfrac{2}{3}\right)^3\right]^2\\ \left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{8}{27}\right)^2\)

TH1: \(x+\dfrac{1}{3}=\dfrac{8}{27}\\ x=\dfrac{8}{27}-\dfrac{1}{3}\\ x=-\dfrac{1}{27}\)

TH2: 

\(x+\dfrac{1}{3}=-\dfrac{8}{27}\\ x=-\dfrac{8}{27}-\dfrac{1}{3}\\ x=-\dfrac{17}{27}\)

Vậy, `x \in {-1/27; -17/27}.`

\(\left(x+\dfrac{1}{3}\right)^2= \left(\dfrac{2}{3}\right)^6\)

\(\Rightarrow x+\dfrac{1}{9}=\dfrac{64}{729}\)

\(\Rightarrow x=\dfrac{64}{729}-\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{64}{729}-\dfrac{81}{729}\)

\(\Rightarrow x=-\dfrac{17}{729}\)

Vậy...