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 .

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

\(\Leftrightarrow3x^2-4x-2x^2-1=0\)

\(\Leftrightarrow x^2-4x-1=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4\\x_1x_2=\dfrac{c}{a}=-1\end{matrix}\right.\)

Ta có :

\(A=x_1^2+x_2^2+3x_1x_2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2+3x_1x_2\)

\(=\left(x_1+x_2\right)^2+x_1x_2\)

\(=4^2-1\)

\(=16-1\)

\(=15\)

Mình phân tích ở trên rồi

Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)

=> x - 1 = 0 => x = 0+1 => x = 1

Hoặc x - 2 = 0 => x = 0+2 => x = 2

Hoặc x - 3 = 0 => x = 0+3 => x = 3

Vậy x = 1 hoặc x = 2 hoặc x = 3

Sửa chút nha : (x - 1)(x² - 5x +6)  = 0 

\(\left(x-1\right)\left(x^2-5x+6\right)=\left(x-1\right)\left(x^2-2x-3x+6\right)\)

                                              \(=\left(x-1\right)\left[x\left(x-2\right)-3\left(x-2\right)\right]\)

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

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

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

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

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

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

______

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

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

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

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

______

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

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

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

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

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)

------------

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

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

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

= (x - y)(xy - 5)

------------

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