Bài 4: Phương trình tích

a: \(\Leftrightarrow3x\left(x-3\right)-x\left(2x-9\right)=0\)

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

=>x=0

b: \(\Leftrightarrow x\cdot\dfrac{1}{2}\left(x-3\right)-\left(x-3\right)\left(\dfrac{3}{2}x-1\right)=0\)

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

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

=>x=3 hoặc x=1

c: \(\Leftrightarrow2x\left(x-5\right)-3\left(x-5\right)=0\)

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

=>x=5 hoặc x=3/2

d: \(\text{Δ}=\left(-3\right)^2-4\cdot3\cdot\left(-3\right)=9+36=45>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3-3\sqrt{5}}{2}\\x_2=\dfrac{3+3\sqrt{5}}{2}\end{matrix}\right.\)

\(\Rightarrow4x+2-3x+6=12-8x-12x\Leftrightarrow x+8=12-20x\)

\(\Leftrightarrow21x=4\Leftrightarrow x=\dfrac{4}{21}\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

\(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\\ \Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}-\dfrac{4\left(3-2x\right)}{12}+\dfrac{12x}{12}=0\\ \Leftrightarrow4x+2-3x+6-12+8x+12x=0\\ \Leftrightarrow21x-4=0\\ \Leftrightarrow x=\dfrac{4}{21}\)

\(5-\left(x-6\right)=4\left(3-2x\right)\\ \Leftrightarrow5-x+6=12-8x\\ \Leftrightarrow11-x=12-8x\\ \Rightarrow11-x-12+8x=0\\ \Leftrightarrow7x-1=0\\ \Leftrightarrow x=\dfrac{1}{7}\)

a: Xét ΔHBA vuông tại H và ΔBEA vuông tại B có 

\(\widehat{HAB}\) chung

Do đó: ΔHBA\(\sim\)ΔBEA
Suy ra: AB/AE=HA/BA

hay \(AB^2=AH\cdot AE\)

b: Xét ΔHBE vuông tại H và ΔHAB vuông tại H có

\(\widehat{HBE}=\widehat{HAB}\)

Do đó: ΔHBE\(\sim\)ΔHAB

Suy ra: HB/HA=HE/HB

hay \(HB^2=HE\cdot HA\)

thi tu lam di ban

a, \(4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

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

c, \(\Rightarrow2x-4=6x+3+3x\Leftrightarrow9x+3=2x-4\Leftrightarrow7x=-7\Leftrightarrow x=-1\)

d, đk : x khác -1 ; 1 

\(\Rightarrow x^2-x+x^2+2x+1=x^2+1\Leftrightarrow x^2+x=0\Leftrightarrow x=0\left(tm\right);x=-1\left(ktm\right)\)

c.\(\dfrac{x-2}{3}=x+\dfrac{1+x}{2}\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)}{6}=\dfrac{6x+3\left(1+x\right)}{6}\)

\(\Leftrightarrow2\left(x-2\right)=6x+3\left(1+x\right)\)

\(\Leftrightarrow2x-4=6x+3+3x\)

\(\Leftrightarrow7x=-7\)

\(\Leftrightarrow x=-1\)

d.\(\dfrac{x}{x+1}+\dfrac{x+1}{x-1}=\dfrac{x^2+1}{x^2-1}\)

\(ĐK:x\ne\pm1\)

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

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

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

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

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

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

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

Câu 3: B/ x = -1.

A. góc B = 70^o.

B

C

\(C/\dfrac{2x}{3}+1=0.\)