a)\(2x^3=x^2+2x-1\)

\(\Rightarrow2x^3-x^2-2x+1=0\)

\(\Rightarrow x^2\left(2x-1\right)-\left(2x-1\right)=0\)

\(\Rightarrow\left(x^2-1\right)\left(2x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\pm1\\x=\frac{1}{2}\end{cases}}\)

b)\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-x\right)\)

\(\Rightarrow\left(3x-1\right)\left(x^2+2\right)-6x\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(x^2+2-6x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\\Delta_{x^2-6x+2=0}=\left(-6\right)^2-4\cdot1\cdot2=28\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x_{2,3}=\frac{6\pm\sqrt{28}}{2}\end{cases}}\)

\(ĐKXĐ:x\ne\pm1\)

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

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

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

\(\Leftrightarrow B=-\left(x^2+1\right).\left(x-1\right)\)

\(\Leftrightarrow B=-x^3+x^2-x+1\)

b) Để B < 0

\(\Leftrightarrow-x^3+x^2-x+1< 0\)

\(\Leftrightarrow-\left(x^2+1\right)\left(x-1\right)< 0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)>0\)

TH1 : \(\hept{\begin{cases}x^2+1>0\left(tm\right)\\x-1>0\end{cases}\Leftrightarrow x>1}\)

TH2 : \(\hept{\begin{cases}x^2+1< 0\left(ktm\right)\\x-1< 0\end{cases}}\Leftrightarrow x\in\varnothing\)

Vậy để \(B< 0\Leftrightarrow x>1\)

c) Khi \(x-4=5\)

\(\Leftrightarrow x=9\)

\(\Leftrightarrow B=-\left(9^3\right)+9^2-9+1\)

\(\Leftrightarrow B=-729+81-9+1\)

\(\Leftrightarrow B=-656\)

Vậy khi \(x-4=5\Leftrightarrow B=-656\)

đag nằm ấm k  có giấy  bút  nhap k tiên tính toán b 

giúp đc k , mình cũng lạnh chứ bộ :3

hđt nâng cao hay  cơ bản

mình cần kiến thức cơ bản thui ! với lại mk mới học lớp 6 !!!