Ta có: \(\widehat{ABE}=\dfrac{\widehat{ABC}}{2}\)

\(\widehat{ACF}=\dfrac{\widehat{ACB}}{2}\)

mà \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)

nên \(\widehat{ABE}=\widehat{ACF}\)

Xét ΔABE và ΔACF có

\(\widehat{ABE}=\widehat{ACF}\)

AB=AC

\(\widehat{BAE}\) chung

Do đó: ΔABE=ΔACF

=>BE=CF

a: Xét ΔABI và ΔACI có

AB=AC

BI=CI

AI chung

Do đó: ΔABI=ΔACI

b: Sửa đề: Chứng minh AC=AE

Ta có: CE//AI

=>\(\widehat{AEC}=\widehat{BAI};\widehat{CAI}=\widehat{ACE}\)

mà \(\widehat{BAI}=\widehat{CAI}\)(ΔABI=ΔACI)

nên \(\widehat{AEC}=\widehat{ACE}\)

=>AC=AE

Số tiền để mua 20 cái kẹo là

\(2000\times20=40000\)( đồng )

Giá mỗi cái bánh là

\(40000\div8=5000\)( đồng )

Đáp số 5000 đồng

\(B=x\left(y^2-1\right)+y\left(x^2-1\right)\)

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

\(=xy\left(x+y\right)-\left(x+y\right)\)

=5xy-5

\(f\left(x\right)=ax^2+bx+c\)

Mà: \(f\left(0\right)=2\) thay `x=0` ta có:

\(\Rightarrow f\left(0\right)=a\cdot0^2+b\cdot0+c=2\Rightarrow c=2\) 

       \(f\left(1\right)=7\) thay `x=1` ta có:

\(\Rightarrow f\left(1\right)=a\cdot1^2+b\cdot1+c=7\Rightarrow a+b+c=7\Rightarrow a+b=5\) (vì `c = 2`) 

\(\Rightarrow a=5-b\) (*) 

      \(f\left(-2\right)=-14\) 

\(\Rightarrow f\left(-2\right)=a\cdot\left(-2\right)^2+b\cdot-2+c=-14\)

\(\Rightarrow4a-2b+c=-14\)

\(\Rightarrow4a-2b=-16\) (vì `c=2`) 

\(\Rightarrow2a-b=-8\) (**) 

Thay (*) vào (**) ta có:

\(2\cdot\left(5-b\right)-b=-8\)

\(\Rightarrow10-2b-b=-8\)

\(\Rightarrow-3b=-18\)

\(\Rightarrow b=6\)

\(\Rightarrow a=5-6=-1\)

Vậy: ... 

Ta có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}\)

\(=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\) 

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y+z}=2\\\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\end{matrix}\right.\) 

Ta có: \(\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\) 

\(\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow\left(x+y+z\right)+1=3x\)

\(\Rightarrow\dfrac{1}{2}+1=3x\)

\(\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)  

\(x+y+z=\dfrac{1}{2}\Rightarrow y+z=0\Rightarrow y=-z\)  

\(\dfrac{x+z+2}{y}=2\Rightarrow\dfrac{\dfrac{1}{2}+z+2}{-z}=2\Rightarrow\dfrac{5}{2}+z=-2z\)

\(\Rightarrow3z=-\dfrac{5}{2}\Rightarrow z=-\dfrac{5}{6}\)

\(\Rightarrow y=-\left(-\dfrac{5}{6}\right)=\dfrac{5}{6}\)

Vậy: \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{5}{6};-\dfrac{5}{6}\right)\)

Ta có:

`4x-5y-6xy+7=0`

`⇒(4x-6xy)-5y+7=0`

`⇒2x(2-3y)-5y+7=0`

`⇒6x(2-3y)-15y+21=0` 

`⇒6x(2-3y)+21-15y=0` 

`⇒6x(2-3y)+(10-15y)+11=0`

`⇒-6x(3y-2)-(15y-10)+11=0`

`⇒-6x(3y-2)-5(3y-2)=-11`

`⇒(3y-2)(6x+5)=11` 

Mà `x,y∈Z` 

Ta có bảng: 

Vậy: ...