\(\dfrac{2A}{2A+16.5}=\dfrac{43,66}{100}\)

=> \(200A=43,66.\left(2A+16.5\right)\)

=> \(200A-87,32A=3492,8\)

=> \(112,68A=3492,8\)

=> A= 31 

Cái đó là tìm ra A là bn ạ

\(\left\{{}\begin{matrix}6u_2+u_5=1\\3u_3+2u_4=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6u_1.q+u_1.q^4=1\\3u_1.q^2+2u_1.q^3=-1\end{matrix}\right.\)

\(\Rightarrow u_1\left(6q+q^4+3q^2+2q^3\right)=0\)

\(\Leftrightarrow q^3+2q^2+3q+6=0\)

\(\Leftrightarrow\left(q+2\right)\left(q^2+3\right)=0\)

\(\Leftrightarrow q=-\text{​​}2\)

\(\Rightarrow u_1=\dfrac{1}{4}\)

\(\Rightarrow u_n=u_1.q^{n-1}=\dfrac{1}{4}.\left(-2\right)^{n-1}=\left(-2\right)^{n-3}\)

Gọi độ dài AB và vận tốc dự kiến lần lượt là x,y

Theo đề, ta có hệ:

\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{10}{3}\\\dfrac{x}{y+5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-10y=0\\x-3y=15\end{matrix}\right.\)

=>x=150 và y=45

\(x=\left(\dfrac{1}{2}\right)^3:\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{3-1}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

1/4

\(\dfrac{1}{4}\)

Ptr có `2` nghiệm phân biệt `<=>\Delta' > 0`

   `=>(m+1)^2-m^2+2m-3 > 0`

`<=>m^2+2m+1-m^2+2m-3 > 0`

`<=>m > 1/2`

`=>` Áp dụng Viét có: `{(x_1+x_2=-b/a=2m+2),(x_1.x_2=c/a=m^2-2m+3):}`

Ta có: `1/[x_1 ^2]-[4x_2]/[x_1]+3x_2 ^2=0`

`=>1-4x_1.x_2+3(x_1.x_2)^2=0`

`<=>1-4(m^2-2m+3)+3(m^2-2m+3)^2=0`

`<=>[(m^2-2m+3=1),(m^2-2m+3=1/3):}`

`<=>[(m^2-2m+2=0(VN)),(m^2-2m+8/3=0(VN)):}`

  `=>` Không có `m` thỏa mãn.

\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

Ok

=4090506-\(\dfrac{3}{2023}\)*2021+2020

=4090506-\(\dfrac{6063}{2023}\)+2020

=\(\dfrac{8279174035}{2023}\) tối giản luôn rồi nhé

\(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne9\right)\)

Để \(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}< 0\) thì 

\(\sqrt{x}-3< 0\) ( do \(\sqrt{x}+3\ge3>0\))

\(\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

Để M là số nguyên âm thì \(\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay \(x\in\left\{0;1;4\right\}\)