Vì \(\hept{\begin{cases}AB\perp BC\\DC\perp BC\end{cases}}\)

\(\Rightarrow AB//CD\) (từ vuông góc đến song song)

\(\Rightarrow\widehat{BAI}+\widehat{AIC}=180^o;\widehat{BAI}=45^o\)

\(\Rightarrow\widehat{AIC}=180^o-45^o=135^o\)

b) Vì \(\hept{\begin{cases}CD\perp DE\\EF\perp DE\end{cases}}\)

\(\Rightarrow CD//EF\) (từ vuông góc đến song song)

Mà \(AB//CD\) 

\(\Rightarrow AB//EF\)

c) Vì \(CD//EF\)

\(\Rightarrow CI//EF\left(I\in DC\right)\)

\(\Rightarrow\widehat{CIF}=\widehat{IFE}\) (2 góc so le trong)

Ta có: \(\widehat{AIC}+\widehat{CIF}=180^o;\widehat{AIC}=135^o\) (2 góc kề bù)

\(\Rightarrow\widehat{CIF}=180^o-135^o=45^o=\widehat{IFE}\)

\(C=-\left|5x+1\right|-4\le-4\)

Dấu ''='' xảy ra khi x = -1/5

Vậy GTLN của C bằng -4 tại x = -1/5 

\(C=-\left|5x+1\right|-4\)

Vì \(-\left|5x+1\right|\le0\forall x\)

\(\Rightarrow-\left|5x+1\right|-4\le-4\forall x\)

\(\Rightarrow C_{max}=-4\Leftrightarrow-\left|5x+1\right|=0\Leftrightarrow x=-\frac{1}{5}\)

Ta có: x−227+x−326+x−425+x−445=0x−227+x−326+x−425+x−445=0

⇔x−2927+x−2926+x−2925+x−295=0⇔x−2927+x−2926+x−2925+x−295=0

⇔x−29=0⇔x−29=0

hay x=29

đúng hong = )

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-44}{5}=0\)

<=> \(\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\frac{x-29}{5}=0\)

<=> \(\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{5}=0\)

<=> \(\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{5}\right)=0\)

<=> x - 29 = 0 (vì \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{5}\ne0\)) 

<=> x = 29

Vậy x = 29 

mình sắp lên lớp 4 rùi

\(\frac{x-20}{9}+\frac{x-21}{10}+\frac{x-26}{15}=-3\)

\(\Leftrightarrow\frac{x-20}{9}+1+\frac{x-21}{10}+1+\frac{x-26}{15}+1=0\)

\(\Leftrightarrow\frac{x-11}{9}+\frac{x-11}{10}+\frac{x-11}{15}=0\)

\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{15}\ne0\right)=0\Leftrightarrow x=11\)

\(A=\left|x-\frac{1}{2}\right|-\frac{3}{4}\)

Vì \(\left|x-\frac{1}{2}\right|\ge0\forall x\)

\(\Rightarrow\left|x-\frac{1}{2}\right|-\frac{3}{4}\ge-\frac{3}{4}\forall x\)

\(\Rightarrow A_{min}=-\frac{3}{4}\Leftrightarrow\left|x-\frac{1}{2}\right|=0\Leftrightarrow x=\frac{1}{2}\)

\(A=\left|x-\frac{1}{2}\right|-\frac{3}{4}\ge-\frac{3}{4}\)

Dấu ''='' xảy ra khi x = 1/2

Vậy GTNN của A bằng -3/4 tại x = 1/2 

hmm......để nghĩ đã nha [[ bối rối ]]

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