\(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\)

=>\(\dfrac{x}{8}=\dfrac{y}{64}=\dfrac{z}{216}\)

=>\(\dfrac{x}{2}=\dfrac{y}{16}=\dfrac{z}{54}\)

=>\(\dfrac{x}{1}=\dfrac{y}{8}=\dfrac{z}{27}\)

Đặt \(\dfrac{x}{1}=\dfrac{y}{8}=\dfrac{z}{27}=k\)

=>x=k; y=8k; z=27k

\(2x^2+2y^2-z^2=1\)

=>\(2k^2+2\cdot\left(8k\right)^2-\left(27k\right)^2=1\)

=>\(2k^2+128k^2-729k^2=1\)

=>\(k^2=-\dfrac{1}{599}\)(vô lý)

Vậy: KHông có bộ số (x;y;z) nào thỏa mãn yêu cầu đề bài

TH1: p=2

\(p^2+8=2^2+8=12\) không là số nguyên tố

=>Loại

TH2: p=3

\(p^2+8=3^2+8=17\) là số nguyên tố

\(p^2+2=3^2+2=11\) là số nguyên tố

=>Nhận

TH3: p=3k+1

\(p^2+8=\left(3k+1\right)^2+8=9k^2+6k+9=3\left(3k^2+2k+3\right)⋮3\)

=>p^2+8 không là số nguyên tố

=>Loại

TH4: p=3k+2

\(p^2+8=\left(3k+2\right)^2+8\)

\(=9k^2+12k+4+8=9k^2+12k+12=3\left(3k^2+4k+4\right)⋮3\)

=>p^2+8 không là số nguyên tố

=>Loại

Vậy: p=3

1 are they studying

2 are lying

3 is singing

4 loves watching

5 to do

6 talking

7 are working

8 are you

9 prefers walking - riding

10 is always forgetting

11 dislikes eating

12 is she - is doing

#\(yGLinh\)

cứu tui

\(-\dfrac{5}{7}\cdot\dfrac{2}{11}+\left(-\dfrac{5}{7}\right)\cdot\dfrac{4}{11}+\dfrac{5}{7}\cdot\left(-\dfrac{5}{11}\right)\\ =-\dfrac{5}{7}\cdot\dfrac{2}{11}+\left(-\dfrac{5}{7}\right)\cdot\dfrac{4}{11}+\left(-\dfrac{5}{7}\right)\cdot\dfrac{5}{11}\\ =\left(-\dfrac{5}{7}\right)\cdot\left(\dfrac{2}{11}+\dfrac{4}{11}+\dfrac{5}{11}\right)\\ =\left(-\dfrac{5}{7}\right)\cdot1=-\dfrac{5}{7}\)

\(\dfrac{5}{7}\cdot\dfrac{6}{13}-\left(-\dfrac{7}{13}\cdot\dfrac{5}{7}\right)+\dfrac{1}{7}:\dfrac{1}{2}\\ =\dfrac{5}{7}\cdot\dfrac{6}{13}-\left(-\dfrac{7}{13}\cdot\dfrac{5}{7}\right)+\dfrac{1}{7}\cdot2\\ =\dfrac{5}{7}\cdot\left[\dfrac{6}{13}-\left(-\dfrac{7}{13}\right)\right]+\dfrac{2}{7}\\ =\dfrac{5}{7}\cdot1+\dfrac{2}{7}=1\)

\(\left(2\dfrac{2}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\\ =\left(\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right)\cdot\left(-\dfrac{17}{3}\right)\\ =\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\cdot\dfrac{-17}{3}\\ =\left(\dfrac{32}{15}\cdot\dfrac{3}{32}\right)\cdot\left(\dfrac{9}{17}\cdot\dfrac{-17}{3}\right)\\ =\dfrac{1}{5}\cdot\left(-3\right)=-\dfrac{3}{5}\)

a) 

\(\dfrac{-5}{11}\cdot\dfrac{7}{15}\cdot\dfrac{11}{-5}\cdot\left(-30\right)\\ =\left(\dfrac{-5}{11}\cdot\dfrac{11}{-5}\right)\cdot\left(\dfrac{7}{15}\cdot-30\right)\\ =1\cdot-14\\ =-14\)

b) 

\(-\dfrac{1}{3}\cdot\left(-\dfrac{15}{19}\right)\cdot\dfrac{38}{45}\\ =-\dfrac{1}{3}\cdot\left(-\dfrac{15}{29}\cdot\dfrac{38}{45}\right)\\ =-\dfrac{1}{3}\cdot\left(-\dfrac{15}{29}\cdot\dfrac{2\cdot19}{3\cdot15}\right)\\ =-\dfrac{1}{3}\cdot-\dfrac{2}{3}\\ =\dfrac{2}{9}\)

đừng đăng linh tinh nhé bạn (Phạm Minh Khuê)
 

9.(\(x-1\))² - (\(x-1\))⁴ = 0

   (\(x-1\))²(9 - (\(x-1\))²) = 0

    \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\9-\left(x-1\right)^2=0\end{matrix}\right.\)

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

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

      \(\left[{}\begin{matrix}x=1\\x=-2\\x=4\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- 2; 1; 4}

uh, mình sẽ làm !

Chào các bạn đã kết bạn với tớ, các bạn có khỏe mạnh không, giờ tớ đang học nè, còn các cậu đang làm gì.