a) \(2x^2-5x+1=0\)

\(\Delta=b^2-4ac\Rightarrow\left(-5\right)^2-4.2.1=17>0\)

Phương trình có 2 nghiệm phân biệt:

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-5\right)+\sqrt{17}}{2.2}=\dfrac{5+\sqrt{17}}{4}\)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-5\right)-\sqrt{17}}{2.2}=\dfrac{5-\sqrt{17}}{4}\)

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b) \(4x^2+4x+1=0\)

\(\Delta=b^2-4ac\Rightarrow4^2-4.4.1=0\)

Vậy phương trình có nghiệm kép:

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c) \(5x^2-x+2=0\)

\(\Delta=b^2-4a\Rightarrow\left(-1\right)^2-4.5.2=-39\)

Vậy phương trình vô nghiệm.

\(a,2x^2-5x+1=0\)

\(\Delta=-b^2-4ac\)

\(\Delta=25-8\)

\(\Delta=17\)

Vậy phương trình có `2` nghiệm phân biệt  :

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+\sqrt{17}}{4} \)

\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-\sqrt{17}}{4}\)

\(b,4x^2+4x+1=0\)

\(\Delta=b^2-4ac\)

\(\Delta=16-16=0\)

Vậy phương trình có nghiệm kép :

\(x=\dfrac{-b}{2a}=-\dfrac{4}{8}=-\dfrac{1}{2}\)

\(c,5x^2-x+2=0\)

\(\Delta=1-40\)

\(\Delta=-39\)

Vậy phương trình vô nghiệm .

\(2x^2+6x-4\left(x+3\right)\)

\(=\left(2x^2+6x\right)-4\left(x+3\right)\)

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

\(=\left(x+3\right)\left(2x+4\right)\)

\(=2\left(x+3\right)\left(x+2\right)\)

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\(xy\left(x-y\right)-5x+5y\)

\(=xy\left(x-y\right)-\left(5x-5y\right)\)

\(=xy\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-5\right)\)

______

\(2x^2+3x-4xy-6y\)

\(=\left(2x^2+3x\right)-\left(4xy+6y\right)\)

\(=x\left(2x+3\right)-2y\left(2x+3\right)\)

\(=\left(x-2y\right)\left(2x+3\right)\)

2x² + 6x - 4(x + 3)

= (2x² + 6x) - 4(x + 3)

= 2x(x + 3) - 4(x + 3)

= (x + 3)(2x - 4)

= 2(x + 3)(x - 2)

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xy(x - y) - 5x + 5y

= xy(x - y) - (5x - 5y)

= xy(x - y) - 5(x - y)

= (x - y)(xy - 5)

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2x² + 3x - 4xy - 6y

= (2x² - 4xy) + (3x - 6y)

= 2x(x - 2y) + 3(x - 2y)

= (x - 2y)(2x + 3)

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

Tham khảo:

Giải pt: \(\sqrt{x-2} \sqrt{4-x}=2x^2-5x-1\) - Hoc24

ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2-5x}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2-3x+2-x^2-2x-2+5x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

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

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

\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)

\(\Leftrightarrow6x=2x^2+4\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow2x^2+4-6x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)

\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)

\(\Leftrightarrow3x^2-13x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)

\(c,\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)

\(\Leftrightarrow\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{2-5x}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2-5x}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-2x-x+2-x^2-2x=2-5x\)

\(\Leftrightarrow-5x+2=2-5x\)

\(\Leftrightarrow-5x+5x=2-2\)

\(\Leftrightarrow0=0\)

=>pt luôn có nghiệm với mọi x.