Question

1,giải các phương trình sau
a,(x^2-x-10).(x^2-x-8)-8=0
b,(x-1).(x+1).(x+3).(x+5)+15=0
c,15x^4-8x^3-14x^2-8x+15+0

Answer 1

Answer 2

Answer 3

Answer 4

a) Ta có: \(\left(x^2-x-10\right)\left(x^2-x-8\right)-8=0\)

\(\Leftrightarrow\left(x^2-x\right)^2-18\left(x^2-x\right)+80-8=0\)

\(\Leftrightarrow\left(x^2-x\right)^2-18\left(x^2-x\right)+72=0\)

\(\Leftrightarrow\left(x^2-x\right)^2-6\left(x^2-x\right)-12\left(x^2-x\right)+72=0\)

\(\Leftrightarrow\left(x^2-x-6\right)\left(x^2-x-12\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)+15=0\)

\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x+3\right)+15=0\)

\(\Leftrightarrow\left(x^2+4x\right)^2+3\left(x^2+4x\right)-5\left(x^2+4x\right)-15+15=0\)

\(\Leftrightarrow\left(x^2+4x\right)^2-2\left(x^2+4x\right)=0\)

\(\Leftrightarrow\left(x^2+4x\right)\left(x^2+4x-2\right)=0\)

\(\Leftrightarrow x\left(x+4\right)\left(x+2-\sqrt{6}\right)\left(x+2+\sqrt{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\sqrt{6}-2\\x=-\sqrt{6}-2\end{matrix}\right.\)