Bài 3:
a: Xét ΔAND vuông tại N và ΔAMB vuông tại M có
\(\widehat{ADN}=\widehat{ABM}\)
Do đó: ΔAND~ΔAMB
=>\(\dfrac{AN}{AM}=\dfrac{AD}{AB}\)
=>\(\dfrac{AM}{AN}=\dfrac{AB}{AD}\)
b: Ta có: \(\dfrac{AM}{AN}=\dfrac{AB}{AD}\)
=>\(\dfrac{AM}{AB}=\dfrac{AN}{AD}\)
=>\(\dfrac{AM}{AB}=\dfrac{AN}{BC}\)
Ta có: ΔBAM vuông tại M
=>\(\widehat{MAB}+\widehat{MBA}=90^0\)
Ta có: ΔAND vuông tại N
=>\(\widehat{NAD}+\widehat{NDA}=90^0\)
\(\widehat{MAN}+\widehat{MAB}+\widehat{NAD}=\widehat{BAD}\)
=>\(\widehat{MAN}+90^0-\widehat{ABM}+90^0-\widehat{ADN}=180^0-\widehat{ABM}\)
=>\(\widehat{MAN}-2\cdot\widehat{B}=-\widehat{B}\)
=>\(\widehat{MAN}=\widehat{ABC}\)
Xét ΔMAN và ΔABC có
\(\dfrac{AM}{AB}=\dfrac{AN}{BC}\)
\(\widehat{MAN}=\widehat{ABC}\)
Do đó: ΔMAN~ΔABC
Bài 1:
a: 3x-2=0
=>3x=2
=>\(x=\dfrac{2}{3}\)
b: \(-\dfrac{1}{3}x+1=\dfrac{2}{3}x-3\)
=>\(-\dfrac{1}{3}x-\dfrac{2}{3}x=-3-1\)
=>-x=-4
=>x=4
c: \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
=>\(2x^2-x+10x-5=2x^2+2x-3x-3\)
=>9x-5=-x-3
=>10x=2
=>\(x=\dfrac{1}{5}\)
d: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
=>\(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
=>6x-3-5x+10=x+7
=>x+7=x+7
=>0x=0(luôn đúng)
=>\(x\in\varnothing\)
e: (5x-10)(2-8x)=0
=>\(\left[{}\begin{matrix}5x-10=0\\2-8x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}5x=10\\8x=2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{4}\end{matrix}\right.\)
f: \(9\left(x+2\right)^2=\left(2x+7\right)^2\)
=>\(\left(3x+6\right)^2=\left(2x+7\right)^2\)
=>\(\left(3x+6\right)^2-\left(2x+7\right)^2=0\)
=>(3x+6-2x-7)(3x+6+2x+7)=0
=>(5x+13)(x-1)=0
=>\(\left[{}\begin{matrix}x=-\dfrac{13}{5}\\x=1\end{matrix}\right.\)
g: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
=>\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
=>\(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16
=>x=4(nhận)
