\(\dfrac{5}{11}+\dfrac{3}{5}+\dfrac{6}{11}\)

\(=\left(\dfrac{5}{11}+\dfrac{6}{11}\right)+\dfrac{3}{5}\)

\(=1+\dfrac{3}{5}\)

\(=\dfrac{5}{5}+\dfrac{3}{5}\)

\(=\dfrac{8}{5}\)

Mệnh đề nào em nhỉ?

Ta có:\(16\equiv1\left(mod15\right)\)

\(\Rightarrow16^n\equiv1^n\equiv1\left(mod15\right)\)

\(\Rightarrow16^n-1\equiv1-1\equiv0\left(mod15\right)\left(đpcm\right)\)

Vậy...

Xét \(\Delta ABD\) và \(\Delta EBD\) có:

\(\widehat{ABD}=\widehat{EBD}\)

\(BD\) chung

\(\widehat{A}=\widehat{E}\)

\(\Rightarrow\Delta ABD=\Delta EBD\)(ch-gn)

\(\Rightarrow DA=DE\)

Vì \(DE\perp BC\Rightarrow DE< DC\)

\(\Rightarrow DA< DC\)

\(\dfrac{9}{14}+\dfrac{3}{19}+\dfrac{5}{14}=\left(\dfrac{9}{14}+\dfrac{5}{14}\right)+\dfrac{3}{19}=1+\dfrac{3}{19}=\dfrac{22}{19}\)

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

Bạn không nên tự hỏi rồi tự trả lời nhé!

\(x^2-4x+m=0\)

\(\Delta'=\left(-2\right)^2-m=4-m\)

Để pt có 2 nghiệm pb thì:\(\Delta'>0\)

\(4-m>0\)

\(m< 4\)

Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m\end{matrix}\right.\)

Ta có:\(x_1^2x_2+x_1x^2_2=10\)

\(x_1x_2\left(x_1+x_2\right)=10\)

\(4m=10\)

\(m=\dfrac{5}{2}\)

Vậy....