O A B C D

Ta có AB//CD (2 đáy của hình thang ABCD)

\(\Rightarrow\frac{OA}{OD}=\frac{OB}{OC}=\frac{AB}{CD}\Rightarrow\frac{OA}{OA+AD}=\frac{OB}{OB+BC}=\frac{AB}{CD}\)

Từ \(\frac{OA}{OA+AD}=\frac{AB}{CD}\Rightarrow\frac{OA}{OA+9}=\frac{12}{30}\Rightarrow AO=6cm\)

Từ \(\frac{OB}{OB+BC}=\frac{AB}{CD}\Rightarrow\frac{OB}{OB+15}=\frac{12}{30}\Rightarrow OB=10cm\)

Ta thấy với \(x=0\)thay vào phương trình ta có:

\(0-8.0+21.0-24.0+9=9\ne0\)

\(\Rightarrow x=0\)không là nghiệm của phương trình

Khi đó, chia cả 2 vế cho \(x^2\)ta được: 

\(x^2-8x+21-24.\frac{1}{x}+\frac{9}{x^2}=0\)

\(\Leftrightarrow\left(x^2+6+\frac{9}{x^2}\right)-\left(8x+24.\frac{1}{x}\right)+15=0\)

\(\Leftrightarrow\left(x+\frac{3}{x}\right)^2-8\left(x+\frac{3}{x}\right)+15=0\)

Đặt \(x+\frac{3}{x}=t\)

\(\Rightarrow t^2-8t+15=0\)\(\Leftrightarrow t^2-3t-5t+15=0\)

\(\Leftrightarrow t\left(t-3\right)-5\left(t-3\right)=0\)\(\Leftrightarrow\left(t-3\right)\left(t-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-3=0\\t-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=3\\t=5\end{cases}}\)

+) Nếu \(t=3\)\(\Rightarrow x+\frac{3}{x}=3\)

\(\Rightarrow\frac{x^2+3}{x}=\frac{3x}{x}\)\(\Leftrightarrow x^2+3=3x\)

\(\Leftrightarrow x^2-3x+3=0\)\(\Leftrightarrow x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{3}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=0\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\)\(\forall x\)\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow\)Phương trình vô nghiệm

+) Nếu \(t=5\)\(\Leftrightarrow x+\frac{3}{x}=5\)\(\Leftrightarrow\frac{x^2+3}{x}=\frac{5x}{x}\)

\(\Leftrightarrow x^2+3=5x\)\(\Leftrightarrow x^2-5x+3=0\)

\(\Leftrightarrow x^2-2.\frac{5}{2}x+\frac{25}{4}-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{13}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\frac{-\sqrt{13}}{2}\\x-\frac{5}{2}=\frac{\sqrt{13}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5-\sqrt{13}}{2}\\x=\frac{5+\sqrt{13}}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{\frac{5-\sqrt{13}}{2};\frac{5+\sqrt{13}}{2}\right\}\)

a, \(A=\left(\frac{3}{x^3+x}-\frac{4}{x^2+1}\right):\frac{1}{x}\)ĐKXĐ : \(x\ne0\)

\(=\left(\frac{3}{x\left(x^2+1\right)}-\frac{4x}{x\left(x^2+1\right)}\right)x=\frac{3-4x}{x\left(x^2+1\right)}.x\)

\(=\frac{3x-4x^2}{x\left(x^2+1\right)}=\frac{x\left(3-4x\right)}{x\left(x^2+1\right)}=\frac{3-4x}{x^2+1}\)

b, Theo bài ra ta có : \(\left|x-2\right|=2\)

\(\Leftrightarrow x-2=\pm2\Leftrightarrow x=4;0\)

Thay x = 0 vào phân thức trên : \(\frac{3-4.0}{0^2+1}=\frac{3}{1}=3\)( ktm vì ĐKXĐ : x khác 0 ) 

Thay x =4 vào phân thức trên : \(\frac{3-4.4}{4^2+1}=\frac{3-16}{16+1}=\frac{-13}{17}\)

Vậy \(A=-\frac{13}{17}\)

a) ĐKXĐ : x³ + x \(\ne0\)

=> x(x² + 1) \(\ne0\)

=> \(\hept{\begin{cases}x\ne0\\x^2+1\ne0\end{cases}}\)

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

\(=\left(\frac{3}{x\left(x^2+1\right)}-\frac{4x}{x\left(x^2+1\right)}\right).x=\frac{\left(3-4x\right).x}{x\left(x^2+1\right)}=\frac{3-4x}{x^2+1}\)

b) Khi |x - 2| = 2

=> \(\orbr{\begin{cases}x-2=2\\x-2=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

Khi x = 0 => A = \(\frac{3-4.0}{0^2+1}=\frac{-1}{1}=-1\)

Khi x = 4 => A = \(\frac{3-4.4}{4^2+1}=\frac{3-16}{16+1}=\frac{-13}{17}\)

a) Đặt \(n_{KMnO_4\left(phan.huy\right)}=x\left(mol\right)\)

\(PTHH:2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\)

                     \(x\)        \(-->\)   \(\frac{x}{2}\)  \(->\frac{x}{2}->\frac{x}{2}\)           (mol)                

Có : \(m_{CR\left(giảm\right)}=m_{Oxi}\)

=> \(\frac{x}{2}\cdot32=400-376\Rightarrow x=1,5\left(mol\right)\)

=> \(m_{KMnO_4\left(phan.huy\right)}=1,5\cdot158=237\left(g\right)\)

b) \(n_{O_2}=\frac{x}{2}=0,75\left(mol\right)\)

=> \(V_{O2}=0,75\cdot22,4=16,8\left(l\right)\)

c) \(m_{KMnO_4\left(spu\right)}=400-237=163\left(g\right)\)

Theo pthh : 

\(n_{K_2MnO_4}=\frac{x}{2}=0,75\left(mol\right)\Rightarrow m_{K_2MnO_4}=147,75\left(g\right)\)

\(n_{MnO_2}=\frac{x}{2}=0,75\left(mol\right)\Rightarrow m_{MnO_2}=65,25\left(g\right)\)

=> \(\hept{\begin{cases}\%m_{KMnO_4}=\frac{163}{376}\cdot100\%=43,35\%\\\%m_{K_2MnO_4}=\frac{147,75}{376}\cdot100\%=39,3\%\\\%m_{MnO_2}=\frac{65,25}{376}\cdot100\%=17,35\%\end{cases}}\)

~Thanks!!!~

Ta có: \(\hept{\begin{cases}x^3-3xy^2=10\\y^3-3x^2y=30\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x^3-3xy^2\right)^2=100\\\left(y^3-3x^2y\right)^2=900\end{cases}}\)

\(\Rightarrow\left(x^3-3xy^2\right)^2+\left(y^3-3x^2y\right)^2=1000\)

\(\Leftrightarrow x^6-6x^4y^2+9x^2y^4+y^6-6x^2y^4+9x^4y^2=1000\)

\(\Leftrightarrow x^6+3x^4y^2+3x^2y^4+y^6=1000\)

\(\Leftrightarrow\left(x^2+y^2\right)^3=1000\)

\(\Rightarrow x^2+y^2=10\)

Có: \(x^3-3xy^2=10\)

=> \(x^6+9x^2y^4-6x^4y^2=100\left(1\right)\)

Có: \(y^3-3yx^2=30\)

=> \(y^6-6y^4x^2+9x^4y^2=900\left(2\right)\)

Lấy (1) + (2) ta được:

=> \(x^6+y^6+3x^2y^4+3x^4y^2=1000\)

=> \(\left(x^2+y^2\right)^3=1000\)

=> \(x^2+y^2=10\)

=> \(p=10.\)