a, Vì AD là pg nên \(\dfrac{AB}{AC}=\dfrac{DB}{DC}=\dfrac{8}{6}=\dfrac{4}{3}\)

b, Theo định lí Pytago tam giác ABC vuông tại A

\(BC=\sqrt{AB^2+AC^2}=10cm\)

Theo tỉ lệ thức ta có \(\dfrac{DC}{AC}=\dfrac{DB}{AB}\Leftrightarrow\dfrac{DC}{6}=\dfrac{DB}{8}\)

\(\Leftrightarrow\dfrac{DC}{6}=\dfrac{6}{8}\Leftrightarrow DC=\dfrac{36}{8}=\dfrac{9}{2}cm\)

Cảm ơn bạn nhiều nha.chúc bn luôn thành công.

a) DB/DC = AB/AC = 8/6 = 4/3

b) BC = \(\sqrt{AB^2+AC^2}\)= \(\sqrt{8^2+6^2}\)= 10 cm

DB/DC = 4/3 => DB = 4/3 DC

DB + DC = BC = 10 cm => 7/3 DC = 10 cm => DC = 30/7 cm

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\le0\end{matrix}\right.\Leftrightarrow x=0\)

Thay x=0 vào pt ta có:

\(\sqrt{x}+1=2\sqrt{-x}\\ \Leftrightarrow\sqrt{0}+1=2\sqrt{-0}\\ \Leftrightarrow1=2\left(vô.lí\right)\)

Vậy pt vô nghiệm

đề sai rồi, ko có \(\sqrt{-x}\)đâu, do \(\sqrt{x}\ge0\)

1827374838100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000999999999999999999999999999999999999999988888888888863636478348565588888858488545<856599889888///884485585A

\(m^2+3m+6=m^2+\dfrac{2.3}{2}m+\dfrac{9}{4}+\dfrac{15}{4}=\left(m+\dfrac{3}{2}\right)^2+\dfrac{15}{4}>0\)(luôn đúng) 

Vậy ta có đpcm 

\(\dfrac{x+1}{98}+\dfrac{x+2}{97}+\dfrac{x+90}{9}+\dfrac{x+84}{15}>-4\\ \Leftrightarrow\left(\dfrac{x+1}{98}+1\right)+\left(\dfrac{x+2}{97}+1\right)+\left(\dfrac{x+90}{9}+1\right)+\left(\dfrac{x+84}{15}+1\right)>0\\ \Leftrightarrow\dfrac{x+99}{98}+\dfrac{x+99}{97}+\dfrac{x+99}{9}+\dfrac{x+99}{15}>0\\ \Leftrightarrow\left(x+99\right)\left(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}\right)>0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}>0\Rightarrow x+99>0\Rightarrow x>-99\)

cho tớ hỏi ông