Bài 4: Phương trình tích

a) \(4x-16=3x\left(x-4\right)\)

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

\(3x\left(x-4\right)-4\left(x-4\right)=0\)

\(\left(x-4\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)

b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\left(đk:x\ne0,2\right)\)

\(\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

\(x^2+2x-x+2=2\)

\(x^2+x=0\)

\(x\left(x+1\right)=0\)

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

a) \(4x-16=3x\left(x-4\right)\)

\(\Leftrightarrow\) \(4\left(x-4\right)=3x\left(x-4\right)\)

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

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

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

TH1 \(\Leftrightarrow3x=0+4\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=4\div3\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

TH2 \(x-4=0\)

 \(\Leftrightarrow\)      \(x=0+4\)

\(\Leftrightarrow x=4\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\dfrac{4}{3}\\4\end{matrix}\right.\)

b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

  \(\Leftrightarrow\) \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{1\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

 \(\Leftrightarrow\) \(x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x+2=2\)

\(\Leftrightarrow x^2+x=2-2\)

\(\Leftrightarrow x^2+x=0\)

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

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

\(\Leftrightarrow\)\(x=0\)

\(\Leftrightarrow\)  \(x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Leftrightarrow x=-1\)

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

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

\(\Leftrightarrow4x^2-20x+25-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)

\(\Leftrightarrow3x^2-24x+21=0\)

\(\Leftrightarrow3x^2-21x-3x+21=0\)

\(\Leftrightarrow3x\left(x-7\right)-3\left(x-7\right)=0\)

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

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

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

\(\Leftrightarrow x^2-x-3x+3=0\)

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

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

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

\(\dfrac{40}{20}\)\(-\dfrac{5(3x+7)}{20}\)\(+\dfrac{4(x+17)}{20}\)=\(\dfrac{0}{20}\)

<=>40-5(3x+7)+4(x+17)=0

<=>40-15x-35+4x+68=0

<=>-15x+4x=-40+35-68

<=>-11x=-73

<=>x=-73:(-11)=\(\dfrac{73}{11}\)

Vậy S={\(\dfrac{73}{11}\)}

bạn đợi mình chút

<=>x(x-1)(x+1)(x+2)=6

<=>(x^2+x)(x^2+x-2)=6

đặt x^2 + x = t

<=>t(t-2)=6

<=>t^2 -2t + 1 = 7

<=>(t-1)^2 = 7

<=> (x^2 + x)^2=7

<=>x^2 + x = căn 7<=> (x+1/2)^2 = căn7 + 5/4

=> x+1/2 = căn(căn7+5/4) = 1,973765769

<=>x=1,473765769

TH2:x+1/2 = -căn(căn7 + 5/4) = -1,973765769

<=>x=-2.473765769

mặt khác (x+1/2)^2=-căn7 + 5/4

=>x+1/2 = căn(-căn7 +5/4) => vô lí 

TH4:x+1/2 = -căn(-căn7 + 5/4) => vô lí

=> TH 3 4 loại 

vậy phương trình có tập nghiệm S = 1,473765769;-2.473765769

a: Khi m=3 thì pt sẽ là 0x+0=0(luôn đúng)

b: Để phương trình có nghiệm duy nhất thì m-3<>0

hay m<>3

Để phương trình có vô số nghiệm thì m-3=0

hay m=3

Tách nhỏ câu hỏi ra bạn

1: =>(x+4)(x-3)=0

=>x=-4 hoặc x=2

2: =>(x-6)(x+1)-2(x+1)=0

=>(x+1)(x-8)=0

=>x=-1 hoặc x=8

3: =>x²-2x-3=0

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

=>x=3 hoặc x=-1

\(1,x^2+x-12=0\\ \Leftrightarrow\left(x^2+4x\right)-\left(3x+12\right)=0\\ \Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ 2,\left(x-6\right)\left(x+1\right)=2\left(x+1\right)\\ \Leftrightarrow\left(x-6\right)\left(x+1\right)-2\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-6-2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)

\(3,x^2-2x+1=4\\ \Leftrightarrow\left(x-1\right)^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-2\\x-1=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ 4,\left(x-3\right)\left(x+3\right)=16\\ \Leftrightarrow x^2-9-16=0\\ \Leftrightarrow x^2-25=0\\ \Leftrightarrow x=\pm5\\ 5,x^2-x-6=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\\ \Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

\(6,\left(2x-1\right).\dfrac{5x-10}{10}+\left(2x-1\right).\dfrac{3x+12}{6}-\left(2x-1\right)x=0\\ \Leftrightarrow\left(2x-1\right).\dfrac{x-2}{2}+\left(2x-1\right).\dfrac{x+4}{2}-\left(2x-1\right)x=0\\ \Leftrightarrow\left(2x-1\right)\left(\dfrac{x-2}{2}+\dfrac{x+4}{2}-x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(\dfrac{x-2}{2}+\dfrac{x+4}{2}-\dfrac{x}{2}\right)=0\\ \Leftrightarrow\left(2x-1\right).\dfrac{x-2+x+4-x}{2}=0\)

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

a: \(\Leftrightarrow\left(x-1+2x+3\right)\left[\left(x^2-2x+1\right)-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]-\left(3x+2\right)\left(9x^2-6x+4\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9-9x^2+6x-4\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(-6x^2+15x+9\right)=0\)

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

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

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

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

b: \(\Leftrightarrow\left(x^2-4x\right)^2+4\left(x^2-4x\right)-2\left(x^2-4x\right)-8=0\)

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

\(\Leftrightarrow\left(x-2\right)^2\cdot\left[\left(x-2\right)^2-6\right]=0\)

hay \(x\in\left\{2;\sqrt{6}+2;-\sqrt{6}+2\right\}\)

a: (2x+1)(3-x)(4-2x)=0

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

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

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

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

=>x=3 hoặc x=-5/2

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

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

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

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

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

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

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

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

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

a: \(\Leftrightarrow2x^3-5x^2-3x=0\)

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

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

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

hay \(x\in\left\{0;3;-\dfrac{1}{2}\right\}\)

b: \(\Leftrightarrow\left(3x-1\right)\left(x^2+x\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+x-7x+10\right)=0\)

=>3x-1=0

hay x=1/3