a,\(11-2x=x-1\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=-4\)

b,\(\text{5(3x+2)=4x+1}\Leftrightarrow15x+10=4x+1\Leftrightarrow15x-4x=1-10\Leftrightarrow11x=-9\Leftrightarrow x=\dfrac{-9}{11}\)

c,\(x^2-4-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x+2\right)\left(x-2\right)-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x-2\right)[\left(x+2\right)-\left(x-5\right)]\Leftrightarrow\left(x-2\right)\left[x+2-x+5\right]\Leftrightarrow\left(x-2\right)7\Leftrightarrow7x-14\)

ko có đáp án nhé . mk nói thật luôn . nếu sai bạn đừng k nhé !

X = 2,340847617

a) 2x²-4x-x+2=0

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

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

=> \(\left[{}\begin{matrix}2x-1=0\\x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

b) 3x²-12x+5x-20=0

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

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

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x³+2x²-x²-2x+2x+4=0

=> x²(x+2)-x(x+2)+2(x+2)=0

=>(x²-x+2)(x+2)=0

=> x=-2( vi x²-x+2>0)

d) x³-x²-4x²+4x+4x-4=0

=> x²(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x²-4x+4)=0

=> \(\left[{}\begin{matrix}x-1=0\\x^2-4x+4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2x²-5x+2=0

⇔2x²-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

⇔\(\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\2x=1\Leftrightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x³+x²+4=0

⇔x³+2x²-x²-2x+2x+4=0

⇔(x³+2x²)-(x²+2x)+(2x+4)=0

⇔x²(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x²-x+2)=0

⇔x+2=0 và x²-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}

\(\sqrt{x^2+2x+1}+\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-3\right)^2}\)

\(=\left|x+1\right|+\left|x-3\right|\)

\(=x+1+x-3\)

\(=2x-2\)

\(=2\left(x-1\right)\)

Thực hiện các phép đổi tương đương , ta đưa ( 1 ) về dạng :

\(\frac{x+4}{2x^2-5x+2}-\frac{x+4}{2x^2-7x+3}=0\)

\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{2x^2-5x+2}-\frac{1}{2x^2-7x+3}\right)=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(1-2x\right)}{\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)}=0\)

\(\Leftrightarrow\left(x+4\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)

Thữ vào mẫu thức : Với \(x=\frac{1}{2}\) thì \(2x^2-5x+2=0\)

Với \(x=-4\) thì \(\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)\ne0\)

Vậy phương trình ( 1 ) là cho nghiệm duy nhất là \(x=-4\)