Bài 7:

a: \(x^3-4x^2-x+4=0\)

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

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>(x-4)(x-1)(x+1)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=1\\ x=-1\end{array}\right.\)

b: \(x^3+4x-5=0\)

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

=>\(\left(x-1\right)\left(x^2+x+5\right)=0\)

=>x-1=0

=>x=1

c: \(x^4+2x^3+5x^2+4x-12=0\)

=>\(x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

=>\(\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

=>\(\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)

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

=>(x-1)(x+2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-2\end{array}\right.\)

d: \(x^3-7x^2+15x-25=0\)

=>\(x^3-5x^2-2x^2+10x+5x-25=0\)

=>\(\left(x-5\right)\left(x^2-2x+5\right)=0\)

=>x-5=0

=>x=5

Bài 5:

a: \(\left(x-5\right)\left(4x-7\right)+x^2-25=0\)

=>(x-5)(4x-7)+(x-5)(x+5)=0

=>(x-5)(4x-7+x+5)=0

=>(x-5)(5x-2)=0

=>\(\left[\begin{array}{l}x-5=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=\frac25\end{array}\right.\)

b: \(\left(x+7\right)\left(3x-5\right)-x^2+49=0\)

=>(x+7)(3x-5)-(x-7)(x+7)=0

=>(x+7)(3x-5-x+7)=0

=>(x+7)(2x+2)=0

=>2(x+1)(x+7)=0

=>(x+1)(x+7)=0

=>\(\left[\begin{array}{l}x+1=0\\ x+7=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-7\end{array}\right.\)

c: \(\left(3x+1\right)^2=\left(4x-5\right)^2\)

=>\(\left(4x-5\right)^2-\left(3x+1\right)^2=0\)

=>(4x-5-3x-1)(4x-5+3x+1)=0

=>(x-6)(7x-4)=0

=>\(\left[\begin{array}{l}x-6=0\\ 7x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=\frac47\end{array}\right.\)

b:

a: \(x^2-11x+30=0\)

=>\(x^2-5x-6x+30=0\)

=>x(x-5)-6(x-5)=0

=>(x-5)(x-6)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\end{array}\right.\)

b: \(10x^2-13x+3=0\)

=>\(10x^2-10x-3x+3=0\)

=>10x(x-1)-3(x-1)=0

=>(x-1)(10x-3)=0

=>\(\left[\begin{array}{l}x-1=0\\ 10x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=\frac{3}{10}\end{array}\right.\)

c: \(3x^2+10x-1=0\)

=>\(x^2+\frac{10}{3}x-\frac13=0\)

=>\(x^2+2\cdot x\cdot\frac53+\frac{25}{9}-\frac{28}{9}=0\)

=>\(\left(x+\frac53\right)^2=\frac{28}{9}\)

=>\(x+\frac53=\pm\frac{2\sqrt7}{3}\)

=>\(x=\pm\frac{2\sqrt7}{3}-\frac53\)

d: \(2x^2-3x+17=0\)

\(\Delta=\left(-3\right)^2-4\cdot2\cdot17=9-8\cdot17=9-136=-127<0\)

=>Phương trình vô nghiệm