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

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

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

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

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

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

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

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

Vậy: \(S=\left\{7;1\right\}\)

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

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

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

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

Vậy: \(S=\left\{1;3\right\}\)

\(d,x\left(2x-7\right)-4x+14=0\)

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

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

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

d: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

e: =>(2x-5-x-2)(2x-5+x+2)=0

=>(x-7)(3x-3)=0

=>x=7 hoặc x=1

f: =>x(x-1)-3(x-1)=0

=>(x-1)(x-3)=0

=>x=1 hoặc x=3

a); b) Do tích = 0 

=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)

=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)

a; *x-1=0 <=>x=1

    *2x+5=0 <=>x=-2,5

    *x²+2=0 <=> ko có x

b; tương tự a

a/ \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)

Vì \(x^2\ge0\Rightarrow x^2+2\ge2>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

Đặt \(y=2x^2-3x-1\), khi đó phương trình trở thành \(y^2-3\left(y-4\right)-16=0\Leftrightarrow y^2-3y-4=0\left(NX:a-b+c=0\right)\)

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

Với \(y=-1\Rightarrow2x^2-3x-1=-1\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{2}\end{matrix}\right.\)

Với \(y=4\Rightarrow2x^2-3x-1=4\Leftrightarrow2x^2-3x-5=0\left(NX:a-b+c=0\right)\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)

KL: Vậy phương trình có nghiệm \(x=0;x=\frac{3}{2};x=-1;x=5\)

a) <=> 3x-2=0 hoặc 4x+5=0

1) 3x-2=0 <=> 3x=2 <=> x=2/3

2) 4x+5=0 <=> 4x=-5 <=> x= -5/4

a) tách ra 2 cái rồi tính mỗi cái

b) phân tích ra ta đc:

 x² - 8x + 16 - x² + 6x -2x + 12=0

sau đó bạn tự giải ra

c) áp dụng hằng đẳng thức ta đc

 (2x+1)(2x-1)=(2x+1)(3x-5)

5(+x)-4=24

8

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

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

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

\(\Delta'=\left(-2\right)^2-3.\left(-8\right)=4+24=28>0.\)

\(\Rightarrow\) Pt có 2 nghiệm phân biệt \(x_1;x_2.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2+2\sqrt{7}}{3}.\\x_2=\dfrac{2-2\sqrt{7}}{3}.\end{matrix}\right.\)