chu vi hình chữ nhật là:

       5 * 4 = 20(cm)

chiều rộng hình chữ nhật đó là:

      (20 - 2) : 2 = 9 (cm)

                    Đáp số : 9cm

vote mik nhaaaaaa!

Giải:

Chu vi hình chữ nhật đó là :

  5 x 4 = 20 ( cm )

Chiều rộng hình chữ nhật đó là :

  ( 20 - 2 ) : 2 = 9 ( cm )

      Đáp số : 9 cm

Học tốt!!!

Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

\(x^2+x=x\left(x+1\right)\)

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

MTC : \(x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

Quy đồng :  

\(\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)

\(\frac{x+1}{x^2+x}=\frac{x+1}{x\left(x+1\right)}=\frac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{x\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{x-1}{x^2+x+1}=\frac{\left(x-1\right)^2\left(x+1\right)x}{x\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)}\)

\(\frac{x}{x^3-1};\frac{x+1}{x^2+x};\frac{x-1}{x^2+x+1}\)

Ta có:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

         \(x^2+x=x\left(x+1\right)\)

        \(x^2+x+1=x^2+x+1\)

\(\Rightarrow MTC=x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

Quy đồng:

\(\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)

\(\frac{x+1}{x^2+x}=\frac{x+1}{x\left(x+1\right)}=\frac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{x\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{x-1}{x^2+x+1}=\frac{\left(x-1\right)^2x\left(x+1\right)}{x\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

\(=\left(-\frac{x+2}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{2-x}{x+2}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{x^2-4x-4-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{-3x^2-4x-4+x^2-4x-4}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{-2x^2-8}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\frac{-2x^2-8}{\left(x-2\right)\left(x+2\right)}.\frac{-x^2\left(x-2\right)}{x\left(x-3\right)}=\frac{2x^4+8x^2}{x\left(x+2\right)\left(x-3\right)}\)

hồn loàng

đặt \(l\left(x\right)=-x^2-2x+1+3m\) dễ thấy \(3m-7\le g\left(x\right)\le3m+1\) (đạo hàm hoặc tư duy) 

Để \(y_{max}=7\) trên \(\left[0;2\right]\) thì : 

TH1 : \(\hept{\begin{cases}3m+1=7\\3m-7>-7\end{cases}}\Leftrightarrow\hept{\begin{cases}m=2\\m>0\end{cases}}\Leftrightarrow m=2\)

\(\hept{\begin{cases}3m+1=-7\\3m-7< 7\end{cases}}\Leftrightarrow\hept{\begin{cases}m=\frac{-8}{3}\\m< \frac{14}{3}\end{cases}}\Leftrightarrow m=\frac{-8}{3}\)

... 

a, \(x+34⋮x+1\) 

Ta có \(x+34=x+1+33\) 

mà \(x+1⋮x+1\)

 để \(x+34⋮x+1\) thì => \(33⋮x+1\) hay  x+1 \(\inƯ\left(33\right)\) 

\(Ư\left(33\right)=\left\{1;3;11;33\right\}\) 

Ta có bảng sau 

Vậy x \(\in\left\{0;2;10;32\right\}\) 

b, \(3x+13⋮x+1\) 

Ta có 3x+13 = 3(x+1) + 10 

mà \(3\left(x+1\right)⋮x+1\) 

để 3x+13\(⋮x+1\)  thì => \(10⋮x+1\) 

hay x+1 \(\in\) Ư(10)

Ư(10) = {1;2;510}

ta có bảng sau 

 Vậy x \(\in\left\{0;1;4;9\right\}\)