a) x^2 - 3x + 2 = 0

\(\Delta=b^2-4ac=\left(-3\right)^2-4.1.2=1\)

=> pt có 2 nghiệm pb

\(x_1=\frac{-\left(-3\right)+1}{2}=2\)

\(x_2=\frac{-\left(-3\right)-1}{2}=1\)

a) Dễ thấy phương trình có a + b + c = 0 

nên pt đã cho có hai nghiệm phân biệt x₁ = 1 ; x₂ = c/a = 2

b) \(\hept{\begin{cases}x+3y=3\left(I\right)\\4x-3y=-18\left(II\right)\end{cases}}\)

Lấy (I) + (II) theo vế => 5x = -15 <=> x = -3

Thay x = -3 vào (I) => -3 + 3y = 3 => y = 2

Vậy pt có nghiệm ( x ; y ) = ( -3 ; 2 )

a, x1 = 1 , x2 = 2

b, x = -3 , y = 2

c, A = 1

d, x = -1 , x= 3

\(\dfrac{x}{x+3}+\dfrac{6}{x-3}=\dfrac{-18}{9-x^2}\)

\(\Leftrightarrow\dfrac{x}{x+3}+\dfrac{6}{x-3}=\dfrac{18}{x^2-9}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm3\)

\(\dfrac{x}{x+3}+\dfrac{6}{x-3}=\dfrac{18}{x^2-9}\)

\(\Leftrightarrow\dfrac{x.\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{6.\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{18}{\left(x+3\right)\left(x-3\right)}\)

\(\Rightarrow x^2-3x+6x+18=18\)

\(\Leftrightarrow x^2-3x+6x=18-18\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow x=0hoặcx+3=0\)

\(\Leftrightarrow x=0\left(tm\right)hoặcx=-3\left(ktm\right)\)

Vậy phương trình có nghiệm là \(x=0\)

Đề đúng không bạn, nghiệm ra xấu lắm

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=18\)

\(\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=18\)

\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)=18\)

Đặt \(x^2-5x+5=a\)

⇒ \(\left(a-1\right)\left(a+1\right)=18\)

⇒ \(a^2-1=18\)

⇒ \(a=\pm\sqrt{19}\)

...

\(\Leftrightarrow\left(x^2-5x+4\right)\left(x^2-5x+6\right)=18\)

Đặt \(x^2-5x+5=t\)

\(\left(t-1\right)\left(t+1\right)=18\)

\(\Leftrightarrow t^2=19\Leftrightarrow t=\pm\sqrt{19}\)

Với t = căn19 \(\Leftrightarrow x^2-5x+5-\sqrt{19}=0\)

\(\Delta=25-4\left(5-\sqrt{19}\right)=5+4\sqrt{19}\)>0

vậy pt có 2 nghiệm pb 

\(x=\dfrac{5\pm\sqrt{5+4\sqrt{19}}}{2}\)

Với t = -căn19 \(\Leftrightarrow x^2-5x+5+\sqrt{19}=0\)

\(\Delta=25-4\left(5+\sqrt{19}\right)=5-4\sqrt{19}\)>0

vậy pt có 2 nghiệm pb 

\(x=\dfrac{5\pm\sqrt{5-4\sqrt{19}}}{2}\)

a. ĐKXĐ:...

\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)

\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}-3+3\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)

\(\Leftrightarrow2\left(\dfrac{x}{2}-\dfrac{3}{x}\right)^2+6=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)

Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=t\Rightarrow2t^2-13t+6=0\Rightarrow\left[{}\begin{matrix}t=6\\t=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{3}{x}=6\\\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

\(\Leftrightarrow x\left(x-1\right)-\dfrac{x-1}{x^2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{x^2}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\)

\(\Leftrightarrow x=1\)

Đặt \(x^2-2x+3=t\Rightarrow2x-x^2+6=9-t\)

Pt trở thành:

\(t\left(9-t\right)=18\Leftrightarrow t^2-9t+18=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2x+3=3\\x^2-2x+3=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=0\\x^2-2x-3=0\end{matrix}\right.\) (bấm máy)