Bài 1
a) \(3x\left(x^2-4x^3+5\right)-\left(3x^2+15x-7\right)=3x^3-12x^4+15x-3x^2-15x+7=-12x^4+3x^3-3x^2+7\)b)\(\left(2x+1\right)^2+\left(5-4x\right)\left(x+3\right)=4x^2+4x+1+\left(5x+15-4x^2-12x\right)=-3x+16\)
Bài 2
a)\(\left(x-1\right)\left(4x-3\right)-4x\left(x-5\right)=5\Rightarrow\left(4x^2-7x+3\right)-\left(4x^2-20x\right)=5\Rightarrow13x=2\Rightarrow x=\dfrac{2}{13}\)
b)\(\left(x-1\right)^2-81=0\Rightarrow\left(x-1-9\right)\left(x-1+9\right)=0\Rightarrow\left[{}\begin{matrix}x-10=0\\x+8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Bài 2:
a: Ta có: \(\left(x-1\right)\left(4x-3\right)-4x\left(x-5\right)=5\)
\(\Leftrightarrow4x^2-3x-4x+3-4x^2+20x=5\)
\(\Leftrightarrow13x=2\)
hay \(x=\dfrac{2}{13}\)
b: Ta có: \(\left(x-1\right)^2-81=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Bài 1:
a: Ta có: \(3x\left(x^2-4x^3+5\right)-\left(3x^2+15x-7\right)\)
\(=-12x^4-3x^3+15x-3x^2-15x+7\)
\(=-12x^4-3x^3-3x^2+7\)
b: Ta có: \(\left(2x+1\right)^2+\left(5-4x\right)\left(x+3\right)\)
\(=4x^2+4x+1+5x+15-4x^2-12x\)
\(=-3x+16\)
