\(a,3x^2+2x-1\Leftrightarrow3x^2-x+3x-1\Leftrightarrow x\left(3x-1\right)+\left(3x-1\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)\)

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Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

b) ( x - 2 )² + ( x - 2 )( x - 3 ) = 0

<=> ( x - 2 )( x - 2 + x - 3 ) = 0

<=> ( x - 2 )( 2x - 5 ) = 0

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

<=> x( x - 2 ) - 3( x - 2 ) = 0

<=> ( x - 2 )( x - 3 ) = 0

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

a) 0,25x+1,5=0

=> x = (0 - 1,5) : 0,25 = -1,5 : 0,25 = -6

Vậy x = -6.

b) 6,36−5,3x=0

=> x = (0 + 6,36) : 5,3 = 6,36 : 5,3 =\(\dfrac{6}{5}=1,2\)
Vậy x = 1,2.

c) 43x−56=12

=> x = \(\left(\dfrac{1}{2}+\dfrac{5}{6}\right)\): \(\dfrac{4}{3}\) = \(\dfrac{4}{3}:\dfrac{4}{3}=1\)

Vậy x = 1.

d) −59x+1=23x−10

=> \(\dfrac{-5}{9}x-\dfrac{2}{3}x=\dfrac{-11}{9}x=-10-1=-11\)

=> \(x=-11:\dfrac{-11}{9}=9\)

Vậy x = 9.

Bài 3:

a) (3x - 2)(4x + 5)=0

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{-5}{4}\end{matrix}\right.\)

vậy phương trình có tập ngiệm S={\(\dfrac{-5}{4};\dfrac{2}{3}\)}

b) 2x(x-3)-5(x-3)=0

\(\Leftrightarrow\left(x-3\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)

vậy phương trình có tập ngiệm S={\(3;\dfrac{5}{2}\)}

c) 2x(x + 3) + 5(x + 3)=0

\(\Leftrightarrow\left(x+3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\2x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{-5}{2}\end{matrix}\right.\)

vậy phương trình có tập ngiệm S={\(-3;\dfrac{-5}{2}\)}

Bài 1

a) 4x+20=0

<=>4x=-20

<=>x= -5

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

<=>2x -3 =3x²-6x+x+2

<=>2x-3-3x²+6x-x-2 = 0

<=>-3x²+7x-5 =0

<=>x=7/6 ( mk cx ko pít đúng hay ko)

a) \(\dfrac{4x-17}{2x^2+1}=0\) ( 1)

ĐKXĐ : \(2x^2+1\ne0\)

⇔ x² \(\ne\dfrac{-1}{2}\) ( luôn đúng )

( 1) ⇔ 4x - 17 = 0

⇔ x = \(\dfrac{17}{4}\)

KL....

b) \(\dfrac{4}{x-2}-x+2=0\)

⇔ \(\dfrac{4-\left(x-2\right)^2}{x-2}=0\left(x\ne2\right)\)

⇔ 4 - ( x² - 4x + 4) = 0

⇔ 4x - x² = 0

⇔ x( 4 - x) = 0

⇔ x = 0 ( TM) hoặc x = 4 ( TM)

KL...

c) x + \(\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\left(x\ne0\right)\)

Đặt : x + \(\dfrac{1}{x}=t\) ⇒ \(x^2+\dfrac{1}{x^2}=t-2\) , ta có :

t = t - 2 ( Vô lý )

⇒ t ∈ ∅

⇒ Phương trình vô nghiệm

d) \(\dfrac{x-3}{x-2}-\dfrac{x-2}{x-4}=3\dfrac{1}{5}\left(x\ne2;x\ne4\right)\)

⇔ ( x - 3)( x - 4) - ( x - 2)² = \(\dfrac{16}{5}\)( x - 2)( x - 4)

⇔ x² - 7x + 12 - x² + 4x - 4 = \(\dfrac{16}{5}\)( x² - 6x + 8)

⇔ \(\dfrac{40-15x}{5}=\dfrac{16x^2-96x+128}{5}\)

giải nốt nha

\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\\ \dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>-3+3\\ \dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\\ \left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\\ x+2954>0\\ x>-2954\)

a.2mx=0 <=> mx=0

•nếu m=0 thì nghiệm đúng với mọi x

•nếu \(m\ne0\) thì nghiệm đúng với x=0

1) \(x^4-6x^3-x^2+54x-72=0\)

\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)

Tự làm nốt...

2) \(x^4-5x^2+4=0\)

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

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

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

Tự làm nốt...

\(x^4-2x^3-6x^2+8x+8=0\)

\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)

...

\(2x^4-13x^3+20x^2-3x-2=0\)

\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)

Bí