Bài 3:
a: ΔABC vuông tại A
=>\(AB^2+AC^2=BC^2\)
=>\(BC=\sqrt{6^2+8^2}=10\left(cm\right)\)
ΔABD vuông tại A
=>\(AB^2+AD^2=BD^2\)
=>\(BD=\sqrt{6^2+4.5^2}=7,5\left(cm\right)\)
b: Xét ΔABC vuông tại A và ΔADB vuông tại A có
\(\dfrac{AB}{AD}=\dfrac{AC}{AB}\left(\dfrac{6}{4,5}=\dfrac{8}{6}=\dfrac{4}{3}\right)\)
Do đó: ΔABC~ΔADB
c: ΔABC~ΔADB
=>\(\widehat{ABC}=\widehat{ADB}\)
Bài 1:
a: 5x-7=13
=>5x=13+7=20
=>\(x=\dfrac{20}{5}=4\)
b: \(\left(x+6\right)\left(3x-2\right)+x^2-36=0\)
=>\(\left(x+6\right)\left(3x-2\right)+\left(x+6\right)\left(x-6\right)=0\)
=>\(\left(x+6\right)\left(3x-2+x-6\right)=0\)
=>(x+6)(4x-8)=0
=>(x-2)(x+6)=0
=>\(\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
c: \(15-8x=9-5x\)
=>-8x+5x=9-15
=>-3x=-6
=>\(x=\dfrac{6}{3}=2\)
d: \(\dfrac{x-7}{3}=\dfrac{2+x}{4}-5\)
=>\(\dfrac{x-7}{3}=\dfrac{x+2-20}{4}=\dfrac{x-18}{4}\)
=>4(x-7)=3(x-18)
=>4x-28=3x-54
=>x=26
e: \(\dfrac{2-x}{2023}+\dfrac{x}{2025}=\dfrac{1-x}{2024}+1\)
=>\(\dfrac{x-2}{2023}-\dfrac{x}{2025}=\dfrac{x-1}{2024}-1\)
=>\(\left(\dfrac{x-2}{2023}-1\right)-\left(\dfrac{x}{2025}-1\right)=\dfrac{x-2025}{2024}\)
=>\(\left(x-2025\right)\cdot\left(\dfrac{1}{2023}-\dfrac{1}{2025}-\dfrac{1}{2024}\right)=0\)
=>x-2025=0
=>x=2025
f: \(x\cdot\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
=>\(\left(x^2+x\right)\left(x^2+x-2\right)=24\)
=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
=>\(\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
mà \(x^2+x+4=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>=\dfrac{15}{4}\forall x\)
nên \(x^2+x-6=0\)
=>(x+3)(x-2)=0
=>\(\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
