a: \(x^2+4x+3=0\)
=>\(x^2+x+3x+3=0\)
=>x(x+1)+3(x+1)=0
=>(x+1)(x+3)=0
=>\(\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
b: \(2x^2-5x+3=0\)
=>\(2x^2-2x-3x+3=0\)
=>2x(x-1)-3(x-1)=0
=>(x-1)(2x-3)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
c: \(x\left(9x-5\right)=6\)
=>\(9x^2-5x-6=0\)
\(\text{Δ}=\left(-5\right)^2-4\cdot9\cdot\left(-6\right)=241>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{5-\sqrt{241}}{2\cdot9}=\dfrac{5-\sqrt{241}}{18}\\x=\dfrac{5+\sqrt{241}}{18}\end{matrix}\right.\)
d: \(x^4-5x^2+4=0\)
=>\(x^4-x^2-4x^2+4=0\)
=>\(x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
=>\(\left(x^2-1\right)\left(x^2-4\right)=0\)
=>\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\\x=-2\end{matrix}\right.\)
e:
ĐKXĐ: x<>0
\(\left(x+\dfrac{1}{x}\right)^2-3\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)-2\left(x+\dfrac{1}{x}\right)+2=0\)
=>\(\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}-1\right)-2\left(x+\dfrac{1}{x}-1\right)=0\)
=>\(\left(x+\dfrac{1}{x}-1\right)\left(x+\dfrac{1}{x}-2\right)=0\)
=>\(\dfrac{x^2+1-x}{x}\cdot\dfrac{x^2+1-2x}{x}=0\)
=>\(\dfrac{\left(x-1\right)^2}{x^2}\cdot\left(x^2-x+1\right)=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
f: \(\left(x-1\right)\cdot x\left(x+1\right)\left(x+2\right)-24=0\)
=>\(\left(x-1\right)\left(x+2\right)\cdot x\left(x+1\right)-24=0\)
=>\(\left(x^2+x-2\right)\left(x^2+x\right)-24=0\)
=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
=>\(\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
mà \(x^2+x+4=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>=\dfrac{15}{4}>0\forall x\)
nên \(x^2+x-6=0\)
=>(x+3)(x-2)=0
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
g: \(x^4-4x^3+6x^2-4x+1=0\)
=>\(x^4-x^3-3x^3+3x^2+3x^2-3x-x+1=0\)
=>\(x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-3x^2+3x-1\right)=0\)
=>\(\left(x-1\right)^4=0\)
=>x-1=0
=>x=1
