M chia hết ch0 420 hay 20

\(H\ge\left|\left(x+2\right)+\left(4-x\right)\right|\)

\(\Rightarrow H\ge2\)

\(\Rightarrow Hmin=2\Leftrightarrow\left|x-2\right|+\left|x-4\right|=2\)

NẾU \(x< 2\):

\(\left|2-x\right|+\left|4-x\right|=2\)

\(\Leftrightarrow2-x+4-x=2\)

\(\Leftrightarrow6-2x=2\Leftrightarrow x=2\left(KTM\right)\)

NẾU :\(2\le x\le4\)

\(\left|x-2\right|+\left|4-x\right|=2\)

\(\Leftrightarrow x-2+4-x=2\left(TM\right)\)

NẾU :\(x>4\)

\(\Leftrightarrow\left(x-2\right)+\left(x-4\right)=2\)

\(\Leftrightarrow2x-6=2\Rightarrow x=4\left(KTM\right)\)

VẬY:\(Hmin=2\)khi\(2\le x\le4\)

\(\left|2x-6\right|=\hept{\begin{cases}2x-6\left(khi2x-6\ge0\right)\\6-2x\left(khi2x-6< 0\right)\end{cases}}\)

\(\left|2x-6\right|=\hept{\begin{cases}2x-6khix\ge3\\6-2xkhix< 3\end{cases}}\)

\(\left|2x-2\right|=\hept{\begin{cases}2x-2khi2x-2\ge0\\2-2xkhi2x-2< 0\end{cases}}\)

\(\left|2x-2\right|=\hept{\begin{cases}2x-2khix\ge1\\2-2xkhix< 1\end{cases}}\)

KHI \(x< 1\):

\(6-2x+2-2x=6\)

\(\Rightarrow-4x+8=6\)

\(\Rightarrow4x=2\Rightarrow x=\frac{1}{2}\)(THỎA MÃN)

KHI \(1\le x< 3\)

\(6-2x+2x-2=6\)

\(\Rightarrow4=6\)9VÔ NGHIỆM)

KHI: \(x\ge3\)

\(\Rightarrow2x-6+2x-2=6\)

\(\Rightarrow4x=14\Rightarrow x=\frac{7}{2}\)(THỎA MÃN)

Ta có:

x² - (3x - 1) - 2(1 - 3x) = 0

=> x² - 3x + 1 - 2 + 6x = 0

=> x² + 3x - 1 = 0

=> (x² + 3x + 9/4) = 13/4

=> (x + 3/2)² = 13/4

=> \(\orbr{\begin{cases}x+\frac{3}{2}=\sqrt{\frac{13}{4}}\\x+\frac{3}{2}=-\sqrt{\frac{13}{4}}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{13}}{4}\\x=\frac{3+\sqrt{13}}{4}\end{cases}}\)

\(x^2\left(3x-1\right)-2\left(1-3x\right)=0\)

\(\Leftrightarrow x^2\left(3x-1\right)+2\left(3x-1\right)=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(3x-1\right)=0\)

Vì \(x^2+2>0\)\(\Rightarrow\)Để \(\left(x^2+2\right)\left(3x-1\right)=0\)thì \(3x-1=0\)\(\Leftrightarrow x=\frac{1}{3}\)

Vậy \(x=\frac{1}{3}\)

a ) Xét \(\Delta ABD\) và \(\Delta EBC\) có : 

           \(AB=BE\) 

            \(\widehat{ABD}=\widehat{EBC}\) ( cùng bằng \(90^o-\widehat{ABC}\) ) 

            \(BD=BC\)

Suy ra \(\Delta ABD=\Delta EBC\left(c.g.c\right)\)

\(\Rightarrow DA=EC\) ( hai cạnh tương ứng ) 

b , Gọi giao điểm của DA với BC và EC theo thứ tự là H và K

Ta có : \(\Delta ABD=\Delta EBC\left(cmt\right)\)

\(\Rightarrow\widehat{ADB}=\widehat{ECB}\) . Do đó \(\widehat{BDH}=\widehat{KCH}\)

Xét \(\Delta DBH\) và \(\Delta CKH\)có :

\(\widehat{BDH}=\widehat{KCH},\widehat{DHB}=\widehat{CHK}\) nên \(\widehat{DBH}=\widehat{CKH}\)

Do \(\widehat{DBH}=90^o\) nên \(\widehat{CKH}=90^o\)

Vậy \(DA\perp EC\)