Bài 4: Phương trình tích

a) \(x^2+x-42=0\Leftrightarrow x^2+7x-6x-42=0\)

\(\Leftrightarrow x\left(x+7\right)-6\left(x+7\right)=0\Leftrightarrow\left(x-6\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-7\end{matrix}\right.\) vậy \(x=6;x=-7\)

b) \(3x^2-10x+3=0\Leftrightarrow3x^2-x-9x+3=0\)

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

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

a: \(\Leftrightarrow\left(x^2-6x+9\right)^2-15\left(x^2-6x+9+1\right)-1=0\)

\(\Rightarrow\left(x^2-6x+9\right)^2-15\left(x^2-6x+9\right)-16=0\)

\(\Leftrightarrow\left(x^2-6x+9-16\right)\left(x^2-6x+9+1\right)=0\)

\(\Leftrightarrow x^2-6x-7=0\)

=>(x-7)(x+1)=0

=>x=7 hoặc x=-1

b: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

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

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

=>x+1=0

hay x=-1

\(a,\left(x-1\right)\left(x^2+3x-2\right)-\left(x^3-1\right)=0\)

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

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

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

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

C₁ : \(b,\left(x^3+x^2\right)+\left(x^2+x\right)=0\)

\(\Rightarrow x^2\left(x+1\right)+x\left(x+1\right)=0\)

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

\(\Rightarrow x\left(x+1\right)^2=0\)

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

C₂ : \(b,\left(x^3+x^2\right)+\left(x^2+x\right)=0\)

\(\Rightarrow x^3+x^2+x^2+x=0\)

\(\Rightarrow x^3+2x^2+x=0\)

\(\Rightarrow x\left(x^2+2x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)^2=0\)

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

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

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

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

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

Vậy phương trình có tập nghiệm \(S=\left\{1;-\dfrac{1}{2}\right\}\)

b) \(\left(x^3+x^2\right)+\left(x^2+x\right)=0\)

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

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

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

Vậy phương trình có tập nghiệm\(S=\left\{-1\right\}\)

\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)

\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}+\dfrac{1}{c\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{bc\left(b-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\dfrac{ac\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{ab}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{b^2c-bc^2-a^{ 2}c+ac^2+a^2b-ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{1}{abc}\)

4.

\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)

\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)

\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)

\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)

Số không đẹp

a)(x - 4)² - 25= 0

<--> ( x - 4)² - 5² = 0

<--> ( x - 4 - 5 )( x - 4 + 5 ) = 0

<--> ( x - 4 - 5 ) = 0 <--> x - 9 = 0 <--> x = 9

hoặc

<--> ( x - 4 + 5 ) = 0 <--> x + 1 = 0 <--> x = -1

b)bài này tương tự bài a

\(a,\left(x-4\right)^2-25=0\)

\(\Rightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)

\(\Rightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

\(b,\left(x-3\right)^2-\left(x+1\right)^2=0\)

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

\(\Rightarrow-4\left(2x-2\right)=0\)

\(\Rightarrow2\left(x-1\right)=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(c,\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)

\(\Rightarrow2x+3=x-1\)

\(\Rightarrow2x-x=-1-3\)

\(\Rightarrow x=-4\)

\(d,\left(3x-7\right)^2-4\left(x+1\right)^2=0\)

\(\Rightarrow\left(3x-7\right)-\left[2\left(x+1\right)\right]^2=0\)

\(\Rightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\)

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

\(\Rightarrow\left(x-9\right)\left(5x-5\right)=0\)

\(\Rightarrow5\left(x-9\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

b, \(\left(x-3\right)^2-\left(x+1\right)^2=0\)

=> \(\left(x-3\right)^2=0\) và \(\left(x+1\right)^2=0\)

=> x-3= 0

=> x= 3

và x+1= 0

=> x=-1

c, \(\left(x^2-4\right)\left(21+3\right)=\left(x^2-4\right)\left(x-1\right)\)

=> \(\dfrac{\left(x^2-4\right)}{\left(x^2-4\right)}=\dfrac{\left(x-1\right)}{21+3}\)

=> \(1=\dfrac{\left(x-1\right)}{21+3}\)

=> x-1= 21+3

=> x-1= 24

=> x= 24+1

=> x=25

\(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\\ \Rightarrow2x^3+3x^2-8x-12=x^3-x^2-4x+4\\ \Rightarrow x^3+4x^2-4x-16=0\\ \Rightarrow\left(x^3-4x\right)+\left(4x^2-16\right)=0\\ \Leftrightarrow x\left(x^2-4\right)+4\left(x^2-4\right)=0\\ \Leftrightarrow\left(x^2-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=-4\end{matrix}\right.\)

\(\left(3x-7\right)^2-4\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\\ \Leftrightarrow\left(5x-5\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)

a)(x²-4)(2x+3)=(x²-4)(x-1)

=>2x+3 = x-1(cả hai đều chia cho x²-4)

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

=>x+4 =0

=> x = -4

Vậy S={-4}

b)(3x-7)²-4(x+1)²=0

=> (3x-7)²-[2(x+1)]²=0

=>[(3x-7)-2(x+1)][(3x-7)+2(x+1)]=0

=>(3x-7-2x-2)(3x-7+2x+2)=0

=>(x-9)(5x-5)=0

=>5(x-9)(x-1)=0

=> x-9=0 và x-1=0

<=> x = 9 và x=1

Vậy S={1;9}