Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

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

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

Vậy x=3/2 hoặc x=-5

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

a: Ta có: \(\left(2x-3\right)\left(x+5\right)=0\)

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

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

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

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

\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)

a, (–31) . (x +7)=0

<=>  x +7 = 0

<=> x = -7

Vậy x \(\in\left\{-7\right\}\)

b, (8 – x) . (x + 13) = 0

<=> \(\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\)

Vậy x \(\in\left\{8;-13\right\}\)

c,(x²– 25) . (3– x )=0

<=> (x - 5) (x + 5) (3 - x) = 0

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

Vậy x \(\in\left\{5;-5;3\right\}\)

d, ( x - 3 ) (x² + 4) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=3\\x^2=-4\end{matrix}\right.\)_{(vô lý)}

Vậy x \(\in\left\{3\right\}\)

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

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

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

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

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

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

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

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

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

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

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

     x²-16-x²-2x = 0

     -16 - 2x = 0

             2x = -16 

               x = -16/2

               x = -8

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

     (3x-1)(x-2)=0

=> x ∈ {1/3 ; 2 }

c) 6x - 12x² = 0

    6x(1-2x) = 0

=> x ∈ {0; 1/2 }

d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha

d/ 4x (3 - ¹/₄ x) + (x -2) ( x+ 2)

câu d bị  sai đề 

mng giúp với

pls gấp quá

a:Ta có: \(\left(x-2\right)\left(x-\dfrac{1}{2}\right)< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 2\)

b: Ta có: \(\left(x+\dfrac{1}{3}\right)\left(x+1\right)>0\)

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

Em hc bảng xét dáu chx ??

a) Ta có: (x-1)(x-4)>0

\(\Leftrightarrow\left[{}\begin{matrix}x-4>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\)

b) Ta có: (x-6)(x-7)<0

\(\Leftrightarrow\left\{{}\begin{matrix}x-6>0\\x-7< 0\end{matrix}\right.\Leftrightarrow6< x< 7\)

c) Ta có: \(\left(x-1\right)\left(x-2\right)\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\le0\end{matrix}\right.\Leftrightarrow1\le x\le2\)

d) Ta có: \(\left(x-2\right)\left(x-\dfrac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2\ge0\\x-\dfrac{2}{3}\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le\dfrac{2}{3}\end{matrix}\right.\)

a) Ta có: (x-1)(x-4)>0

⇔[x−4>0x−1<0⇔[x>4x<1⇔[x−4>0x−1<0⇔[x>4x<1

b) Ta có: (x-6)(x-7)<0

⇔{x−6>0x−7<0⇔6<x<7⇔{x−6>0x−7<0⇔6<x<7

c) Ta có: (x−1)(x−2)≤0(x−1)(x−2)≤0

⇔{x−1≥0x−2≤0⇔1≤x≤2⇔{x−1≥0x−2≤0⇔1≤x≤2

d) Ta có: (x−2)(x−23)≥0(x−2)(x−23)≥0

⇔⎡⎣x−2≥0x−23≤0⇔⎡⎣x≥2x≤23