Vì \(\left|x+3\right|\ge0\) và \(\left|x-7\right|\ge0\)

\(\Rightarrow\left|x+3\right|+\left|x-7\right|\ge0\)

Dấu = khi : \(\left\{{}\begin{matrix}\left|x+3\right|=0\\\left|x-7\right|=0\end{matrix}\right.\) \(\)

\(\Rightarrow\left\{{}\begin{matrix}x=0-3\\x=0+7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

\(\Rightarrow\) GTNN của \(\left|x+3\right|+\left|x-7\right|\) = 0 khi \(x\in\left\{-3,7\right\}\)

Vì \(\left\{\left|x+1\right|,\left|x+5\right|,\left|x+8\right|,\left|x+10\right|\right\}\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+5\right|+\left|x+8\right|+\left|x+10\right|\ge0\)

Dấu = khi \(\left\{{}\begin{matrix}\left|x+1\right|=0\\\left|x+5\right|=0\\\left|x+8\right|=0\\\left|x+10\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-5\\x=-8\\x=-10\end{matrix}\right.\)

\(\Rightarrow\) GTNN của \(\left|x+1\right|+\left|x+5\right|+\left|x+8\right|+\left|x+10\right|=0\)

khi \(x\in\left\{-1,-5,-8,-10\right\}\)

2

bạn ghi sai đề nên mình sẽ chữa lại đề theo 2 cách để bạn lựa chọn

C1 Tìm GTNN của : \(\left|x+5\right|+\left|x-3\right|\)

Vì \(\left|x+5\right|\ge0,\left|x-3\right|\ge0\)

\(\Rightarrow\) \(\left|x+5\right|-\left|x-3\right|\ge0\)

Dấu = khi \(\left\{{}\begin{matrix}\left|x+5\right|=0\\\left|x-3\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

\(\Rightarrow\) GTNN của \(\left|x+5\right|+\left|x-3\right|\) = 0 khi \(x\in\left\{-5,3\right\}\)

C2 Tìm GTNN của \(\left|x+5\right|+\left|x-3\right|\)

Vì \(\left|x+5\right|\ge0,\left|x-3\right|\ge0\)

\(\Rightarrow\left|x+5\right|+\left|x-3\right|\ge0\)

Dấu = khi \(\left\{{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

\(\Rightarrow\) GTNN của \(\left|x+5\right|+\left|x-3\right|\) = 0 khi \(x\in\left\{-5,3\right\}\)

Ta có: \(10A=10.\left(\frac{10^{2014}+1}{10^{2015}+1}\right)=\frac{10^{2015}+10}{10^{2015}+1}=\frac{10^{2015}+1+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)

\(10B=10.\left(\frac{10^{2015}+1}{10^{2016}+1}\right)=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)

Vì 1 = 1; 9 = 9 ta so sánh mẫu:

Ta có: 10²⁰¹⁵ < 10²⁰¹⁶ => 10²⁰¹⁵+1 < 10²⁰¹⁶+1

=> \(1+\frac{9}{10^{2015}+1}>1+\frac{9}{10^{2016}+1}\)

=> 10A > 10B

=> A > B.

Xét \(\Delta BAE\) và \(\Delta BHE\) có :

BE : cạnh chung

\(\widehat{ABE}=\widehat{HBE}\) (gt)

Ta có : \(\widehat{ABE}+\widehat{BAE}+\widehat{AEB}=\widehat{HBE}+\widehat{BHE}+\widehat{HEB}\)

\(\Rightarrow\widehat{BEA}=\widehat{BEH}\)

\(\Rightarrow\Delta BAE=\Delta BHE\) (g . c . g)

\(\Rightarrow\) AB = BH

Vì \(\Delta BAE\) và \(\Delta BHE\) là tam giác vuông .

\(\Rightarrow\widehat{ABE}+\widehat{AEB}=\widehat{EBH}+\widehat{HEB}\)

\(\Rightarrow\widehat{AEB}=\widehat{HEB}=120^0\)

\(\Rightarrow\widehat{AEB}=\)\(\dfrac{1}{2}\)\(\times120^0=60^0\)

Mà \(\widehat{ABE}+\widehat{BAE}+\widehat{AEB}=180^0\)

\(\Rightarrow\widehat{ABE}=180^0-90^0-60^0=30^0\)

tưởng ngon ăn dùng cô-si ai dè @@

\(\frac{3}{5}=\frac{6}{10}=\frac{9}{15}=\frac{12}{20}\)

Ta có : \(\widehat{ADE}=\widehat{AED}\) (\(\Delta EAB=\Delta DAC\))

Mà \(\widehat{DAE}+\widehat{DEA}+\widehat{EDA}\) = 180 độ

\(\Rightarrow\) 60 + \(\widehat{DEA}+\widehat{EDA}\) =180 độ

\(\Rightarrow\widehat{DEA}+\widehat{EDA}\) = 120 độ

\(\Rightarrow\) \(\widehat{DEA}=\widehat{EDA}=\) \(\dfrac{1}{2}\times120\) = 60 độ