a,

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

b,

\(\left(0,2x-3\right)\left(0,5x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,2x=3\\0,5x=8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\x=16\end{matrix}\right.\)

c,

\(2x\left(x-6\right)+3\left(x-6\right)=0\\ \Leftrightarrow\left(2x+3\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1,5\\x=6\end{matrix}\right.\) (mình skip bớt cho đỡ lằng nhằng nhé :>)

d,

\(\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\\ \Leftrightarrow6\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Chúc bạn học tốt nha

a, x-2=0\(\Leftrightarrow\) x=2

2x-5=0\(\Leftrightarrow\)2x=5\(\Leftrightarrow\)x=\(\dfrac{5}{2}\)

S=\(\left\{\dfrac{5}{2};2\right\}\)

b, 0.2x-3=0\(\Leftrightarrow\)0.2x=3\(\Leftrightarrow\)x=\(\dfrac{3}{0.2}\)

s=\(\left\{\dfrac{3}{0.2}\right\}\)

c, \(\Leftrightarrow\)(x-6)(2x+3)=0

\(\Leftrightarrow\)x-6=0\(\Leftrightarrow\)x=6

2x+3=0\(\Leftrightarrow\)2x=-3\(\Leftrightarrow\)x=\(\dfrac{-3}{2}\)

S=\(\left\{\dfrac{-3}{2};-3\right\}\)

D \(\Leftrightarrow\)x-1=0\(\Leftrightarrow\)x=1

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

3x-9=0\(\Leftrightarrow\)3x=9\(\Leftrightarrow\)x=3

a) (x-2)(2x-5)=0

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

Vậy phương trình có tập nghiệm duy nhất S={2; 2,5}

b) (0,2x-3)(0,5x-8)=0

\(\Leftrightarrow\left[{}\begin{matrix}0,2x-3=0\\0,5x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=16\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S={15;16}

c)2x(x-6)+3(x-6)=0

<=> (x-6)(2x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1,5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S={6;-1,5}

d) (x-1)(2x-4)(3x-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-4=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S={1;2;3}

*vn:vô nghiệm.

a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

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

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

a) Ta có: (5x-1)(x-3)<0

nên 5x-1 và x-3 trái dấu

Trường hợp 1:

\(\left\{{}\begin{matrix}5x-1>0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{5}< x< 3\)

Trường hợp 2:

\(\left\{{}\begin{matrix}5x-1< 0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>3\end{matrix}\right.\Leftrightarrow loại\)

Vậy: S={x|\(\dfrac{1}{5}< x< 3\)}

a: =>x-2=0 hoặc x+3=0

=>x=2 hoặc x=-3

b:=>x-7=0 hoặc x+2=0

=>x=7 hoặc x=-2

c: =>4x+2=0 hoặc 3x-4=0

=>x=4/3 hoặc x=-1/2

d: =>2x+1=0 hoặc x-3=0

=>x=3 hoặc x=-1/2

a)

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

`<=> x-2=0` hoặc `x+3=0`

`<=>x=2` hoặc `x=-3`

b)

`(x-7)(2+x)=0`

`<=>x-7=0` hoặc `2+x=0`

`<=>x=7` hoặc `x=-2`

c)

`(4x+2)(3x-4)=0`

`<=>4x+2=0` hoặc `3x-4=0`

`<=>x=-1/2` hoặc `x=4/3`

d)

`(2x+1)(x-3)=0`

`<=>2x+1=0` hoặc `x-3=0`

`<=>x=-1/2` hoặc `x=3`

e)

`(0,1x-3)(x+0,5)=0`

`<=>0,1x-3=0` hoặc `x+0,5=0`

`<=>x=30` hoặc `x=-0,5`

f)

`(0,2x-0,4)(0,1x+0,7)=0`

`<=>0,2x-0,4=0` hoặc `0,1x+0,7=0`

`<=>x=2` hoặc `x=-7`

a. 2x\(^2\)-8=0

2x\(^2\)=8

x\(^2\)=4

x=2

b.3x\(^3\)-5x=0

x(3x\(^2\)-5)=0

\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)

c.x+3x-4=0\(^{\left(\cdot\right)}\)

đặt t=x\(^2\) (t>0)

ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)

thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm

t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4

khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1

khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2

vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2

d)3x\(^2\)+6x-9=0

thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm

x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)

e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)  (ĐK: x#5; x#2 )

⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)

⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0

⇔-7x\(^2\) - 6x + 46=0

Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0

\(\sqrt{\Delta'}=\sqrt{62}\)

vậy pt có 2 nghiệm phân biệt

x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)

x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)

vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......

câu g làm tương tự câu c

Bài 3: 

b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)

hay \(x\in\left\{0;-1\right\}\)

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

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

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

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

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

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

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

\(b,\left(x-2\right)\left(x-5\right)=0\)

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

\(c,\left(x+3\right)\left(x-5\right)=0\)

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

\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)

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

\(e,\left(x-4\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

\(f,\left(2x-1\right)\left(3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

`a,(x-1)(x+2)=0`

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

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

`b,(x -2)(x -5)=0`

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

`c,(x +3)(x -5)=0`

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

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

`d,(x + 1/2)(4x + 4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)

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

`e,(x -4)(5x -10)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

`f,(2x -1)(3x +6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)

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

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

`g,(2,3x -6,9)(0,1x -2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)

a.(x - 1)(x + 2)= 0

<=> x-1=0 hoặc x+2=0

<=> x=1 hoặc x=-2

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

<=> x-2=0 hoặc x-5=0

<=> x=2 hoặc x=5

c.(x +3)(x -5)=0

<=> x+3=0 hoặc x-5=0

<=> x=-3 hoặc x=5

d.(x + 1/2)(4x + 4)=0

<=> x+1/2=0 hoặc 4x+4=0

<=> x=-1/2 hoặc x=-1

e.(x -4)(5x -10)=0

<=> x-4=0 hoặc 5x-10=0

<=> x=4 hoặc x=2

f.(2x -1)(3x +6)=0

<=> 2x-1=0 hoặc 3x+6=0

<=> x=1/2 hoặc x=-2

g.(2,3x -6,9)(0,1x -2)=0

<=> 2,3x-6,9=0 hoặc 0,1x-2=0

<=> x=3 hoặc x=20

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

a) (x - 7)(2x + 8) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy: S = {7; -4}

b) Tương tự câu a

c)  (x - 1)(2x + 7)(x² + 2) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)

Mà: x² + 2 > 0 với mọi x

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)

d) (2x - 1)(x + 8)(x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)

a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{7;-4\right\}\)

b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)

c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)

a)(x-7)(2x+8)=0

⇔x-7=0 hoặc 2x+8=0

1.x-7=0⇔x=7

2.2x+8=0⇔2x=-8⇔x=-4

phương trình có 1 nghiệm x=7 và x=-4

b)(3x+1)(5x-2)=0

⇔3x+1=0 hoặc 5x-2=0

1.3x+1=0⇔3x=-1⇔x=-1/3

2.5x-2=0⇔5x=2⇔x=5/2

phương trình có 2 nghiệm x=-1/3 và x=5/2