a: Xét ΔABD vuông tại D và ΔACE vuông tại E có

AB=AC
\(\widehat{BAD}\) chung

Do đó: ΔABD=ΔACE

Suy ra: BD=CE

b: Xét ΔEBC vuông tại E và ΔDCB vuông tại D có

BC chung

EC=DB

Do đó:ΔEBC=ΔDCB

Suy ra: \(\widehat{KCB}=\widehat{KBC}\)

hay ΔKBC cân tại K

d: Xét ΔABK và ΔACK có

AB=AC
BK=CK

AK chung

Do đó: ΔABK=ΔACK

Suy ra: \(\widehat{BAK}=\widehat{CAK}\)

hay AK là tia phân giác của góc BAC

\(x^2+4x+y^2-2xy+x^2+4=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)=0\)

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

vì \(\left(x-y\right)^2\ge0;\left(x+2\right)^2\ge0\)nên

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-y=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=-2\end{cases}\Rightarrow}x=y=-2}\)

\(Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)

\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+xy+yz+zx\right)}+\sqrt{6\left(y^2+xy+yz+zx\right)}+\sqrt{z^2+xy+yz+zx}}\)

\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{3\left(x+y\right).2\left(x+z\right)}+\sqrt{3\left(y+x\right).2\left(y+z\right)}+\sqrt{\left(z+x\right).\left(z+y\right)}}\)

\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(y+x\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}}\)

\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{9x+9y+6z}{2}}=\frac{2}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)và  \(z=2\)

\(6^{2.x+5}=216\)

\(=>6^{2.x+5}=6^3\)

=>2.x+5=3

=>2.x=3-5

=>2.x=-2

=>x=-2:2

=>x=-1

`6^(2x+5)=216`

`=> 6^(2x+5)=6^3`

`=>2x+5=3`

`=>2x=3-5`

`=>2x=-2`

`=>x=-2:2`

`=>x=-1`

`6^(2x+5)=216`

`6^(2x+5)=6^3`

`2x+5=3`

`2x=3-5`

`2x=-2`

`x=-2:2`

`x=-1`

\(\left(-12\right).65+7.12-8.\left(-12\right)\)

\(=-12\left(65-7-8\right)\)

\(=-12.50\)

\(=-600\)

\(2^{x+3}.2=2^2.3+52\)

\(=>2^{x+3}.2=64\)

\(=>2^{x+3}=64:2\)

\(=>2^{x+3}=32\)

\(=>2^{x+3}=2^5\)

=>x+3=5

=>x=5-3

=>x=2

Vậy ...........

 2^{x + 3} . 2 = 2² . 3 + 52

 2^{x + 3} . 2 = 4 . 3 + 52

 2^{x + 3} . 2 = 12 + 52

 2^{x + 3} . 2 = 64

     2^{x + 3}  = 64 : 2

     2^{x + 3}  = 32

     2^{x + 3} = 2⁵

     x + 3 = 5

           x = 5 - 3

           x = 2

Vậy x = 2

A max khi x+3=1

=>x=-2

(n+3) ⋮ (2n-1)

=> 2.(n+3)⋮2n-1

=> 2n+6 ⋮ 2n-1

=> (2n-1)+7⋮2n-1

mà 2n-1⋮2n-1

=> 7⋮2n-1

=>2n-1∈Ư(7)={1;7}

=>2n∈{2;8}

=>n∈{1;4}

Vậy n∈{1;4}

\(=\dfrac{1}{3}:\dfrac{7\cdot\left(5+3\right)}{18\cdot\left(53-11\right)}\) 

\(=\dfrac{1}{3}:\dfrac{7\cdot8}{18\cdot42}\) 

\(=\dfrac{1}{3}:\dfrac{7\cdot2\cdot2\cdot2}{3\cdot3\cdot2\cdot2\cdot3\cdot7}\)

\(=\dfrac{1}{3}:\dfrac{2}{3\cdot3\cdot3}\)

\(=\dfrac{1}{3}\cdot\dfrac{27}{2}\)

\(=\dfrac{9}{2}\)