22/ \(\omega A=8\pi\)

\(A^2=x^2+\dfrac{v^2}{\omega^2}\Leftrightarrow A^2=3,2^2+\dfrac{\left(4,8\pi\right)^2}{\omega^2}\)

\(\Leftrightarrow\omega^2A^2=3,2^2\omega^2+23,04\pi^2\Leftrightarrow64\pi^2=3,2^2.\omega^2+23,04\pi^2\Leftrightarrow\omega=2\pi\left(rad/s\right)\)

\(\Rightarrow f=\dfrac{\omega}{2\pi}=\dfrac{2\pi}{2\pi}=1\left(Hz\right)\Rightarrow D.1Hz\)

23/ \(\omega A=20;\omega^2A=80\Rightarrow\left\{{}\begin{matrix}\omega=4\left(rad/s\right)\\A=5cm\end{matrix}\right.\)

\(\Rightarrow v=\omega\sqrt{A^2-x^2}=4.\sqrt{5^2-4^2}=12\left(cm/s\right)\Rightarrow A.12cm/s\)

Câu 16: A

Câu 14: C

Câu 12: A

\(-20x+10x^2-5x^2-10x-5x^2-15x=0\)

\(-45x=0\Leftrightarrow x=0\)

\(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\\ \Rightarrow5x\left(x+3\right)+5x\left(x+2\right)+10x\left(2-x\right)=0\\ \Rightarrow5x\left(x+3+x+2+4-2x\right)=0\\ \Rightarrow5x.9=0\\ \Rightarrow45x=0\\ \Rightarrow x=0\)

−10x(2−x)−5x(x+2)=5x(x+3)⇒5x(x+3)+5x(x+2)+10x(2−x)=0⇒5x(x+3+x+2+4−2x)=0⇒5x.9=0⇒45x=0⇒x=0

x/-30+1,75=-5/9:4/3

=>x/-30+7/4=-5/12

=>x/-30=-13/6

=>x.6=-30.-13

=>x.6=390

=>x=65

vậy x=65.

bạn viết rõ ra phân số ra nha!

= (....) . ( 2008 . 2010 . 10001 - 2008 . 10001 . 2010 )

= (....) . 0

= 0

             k                   mik                  nha!

(1+3+5+... 2009).(2008.20102010-20082008.2010)

= ( 1 + 3 + 5 + ... + 2009 ) . ( 2008 . 2010 . 10001 - 2008 . 10001 . 2010 )

= ( 1 + 3 + 5 + ... + 2009 ) . 0

= 0

Bạn hãy tính vế (2008 . 20102010 - 20082008.2010) sẽ ra bằng 0 rồi kết quả bằng 0

Nhớ chọn cho mk nha!!!

1.1

Pt có 2 nghiệm trái dấu và tổng 2 nghiệm bằng -3 khi:

\(\left\{{}\begin{matrix}ac< 0\\x_1+x_2=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(m+2\right)< 0\\\dfrac{2m+1}{m+2}=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< -2\\m=-\dfrac{7}{5}\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại m thỏa mãn

b.

Pt có nghiệm kép khi:

\(\left\{{}\begin{matrix}m+2\ne0\\\Delta=\left(2m+1\right)^2-8\left(m+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne-2\\4m^2-4m-15=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{5}{2}\\m=-\dfrac{3}{2}\end{matrix}\right.\)

a: Ta có: \(P=A\cdot B\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)

Để P nguyên thì \(\sqrt{x}+1\in\left\{1;5\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;4\right\}\)

hay \(x\in\left\{0;16\right\}\)

e) \(15,6\times7,5+15,6\times2,5\)
\(\Rightarrow15,6\times\left(7,5+2,5\right)\)
\(\Rightarrow15,6\times10\)
\(\Rightarrow156\)

= 156